All questions of Rotational Dynamics for Grade 9 Exam
The centre of mass of a body is locateda)outside the systemb)inside or outside the systemc)inside the systemd)at the centre of systemCorrect answer is option 'B'. Can you explain this answer?
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Suresh Iyer answered |
The centre of mass of a body can lie within or outside the body.
Example
(i) Centre of mass of a uniform rod lies at its geometrical centre which lies within the rod
(ii) Centre of mass of a uniform ring lies at its geometrical centre which lies outside the ring.
Example
(i) Centre of mass of a uniform rod lies at its geometrical centre which lies within the rod
(ii) Centre of mass of a uniform ring lies at its geometrical centre which lies outside the ring.
Can you explain the answer of this question below:Two blocks of the masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block. The velocity of centre of mass is
- A:
10 m/s
- B:
5 m/s
- C:
15 m/s
- D:
20 m/s
The answer is a.
Two blocks of the masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block. The velocity of centre of mass is
10 m/s
5 m/s
15 m/s
20 m/s
|
EduRev JEE answered |
Velocity of heavier block (v2) = 14m/s
Velocity of lighter block (v1) = 0m/s
Velocity of centre of mass,
Velocity of lighter block (v1) = 0m/s
Velocity of centre of mass,
A mass m is moving with a constant velocity along a line parallel to the x-axis, away from the origin. Its angular momentum with respect to the origin
- a)Is zero
- b)Remains constant
- c)Goes on increasing
- d)Goes on decreasing
Correct answer is option 'B'. Can you explain this answer?
A mass m is moving with a constant velocity along a line parallel to the x-axis, away from the origin. Its angular momentum with respect to the origin
a)
Is zero
b)
Remains constant
c)
Goes on increasing
d)
Goes on decreasing
|
Krishna Iyer answered |
Angular momentum (L) is defined as the distance of the object from a rotation axis multiplied by the linear momentum
L = mv×y
L = mv×y
As the particle moves, m; v; and y, all remain unchanged at any point of time
⇒ L = constant
⇒ L = constant
There are two objects of masses 1 kg and 2 kg located at (1, 2) and (-1, 3) respectively. The coordinates of the centre of mass are- a)( 2, -1 )
- b)( 8/3 ,-1/3 )
- c)( -1/3 , 8/3 )
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
There are two objects of masses 1 kg and 2 kg located at (1, 2) and (-1, 3) respectively. The coordinates of the centre of mass are
a)
( 2, -1 )
b)
( 8/3 ,-1/3 )
c)
( -1/3 , 8/3 )
d)
none of these
|
Riya Banerjee answered |
Body A has mass of 1kg and location (1,2)
Body B has mass of 2kg and location (-1,3)
Body B has mass of 2kg and location (-1,3)
mcxc = m1x1 + m2x2
(1+2) xc = (1 * 1) + (2 * -1)
xc = -1/3
(1+2) xc = (1 * 1) + (2 * -1)
xc = -1/3
Similarly,
mcyc = m1y1 + m2y2
(1 + 2) yc = (1 * 2) + (2 * 3)
yc= 8/3
mcyc = m1y1 + m2y2
(1 + 2) yc = (1 * 2) + (2 * 3)
yc= 8/3
Hence, the coordinates of the center of mass are (-1/3, 8/3).
The moment of inertia of two spheres of equal masses is equal. If one of the spheres is solid of radius 
m and the other is a hollow sphere. What is the radius of the hollow sphere?- a)5 m
- b)√3 m
- c)3√3 m
- d)3 m
Correct answer is option 'C'. Can you explain this answer?
The moment of inertia of two spheres of equal masses is equal. If one of the spheres is solid of radius m and the other is a hollow sphere. What is the radius of the hollow sphere?
m and the other is a hollow sphere. What is the radius of the hollow sphere?a)
5 m
b)
√3 m
c)
3√3 m
d)
3 m
|
Sushil Kumar answered |
Moment of inertia of solid sphere Is= 2/5MR2
moment of inertia of hollow sphere Ih =2/3MR2
given mass of solid sphere =√45 kg.
Is=Ih
2MR2/5=2MR2/3
given their masses are equal 2 (√45)2/5= 2 R2/3
45/5=R2/3
9=R2/3
9×3=R2
27=R2
√27=R
√3×9=R
3√3 m=R.
moment of inertia of hollow sphere Ih =2/3MR2
given mass of solid sphere =√45 kg.
Is=Ih
2MR2/5=2MR2/3
given their masses are equal 2 (√45)2/5= 2 R2/3
45/5=R2/3
9=R2/3
9×3=R2
27=R2
√27=R
√3×9=R
3√3 m=R.
A right triangular plate ABC of mass m is free to rotate in the vertical plane about a fixed horizontal axis through A. It is supported by a string such that the side AB is horizontal. The reaction at the support A is : 
- a)
- b)
- c)
- d)mg
Correct answer is option 'B'. Can you explain this answer?
A right triangular plate ABC of mass m is free to rotate in the vertical plane about a fixed horizontal axis through A. It is supported by a string such that the side AB is horizontal. The reaction at the support A is :
a)
b)
c)
d)
mg
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Crafty Classes answered |
The distance of Centre Of Mass of the given right angled triangle is 2L/3 along BA and L/3 along AC from the point B.
Force of magnitude mg is acting downwards at its COM.
Moment balance around B gives:
mg(2L/3)−FA(L)=0
(Moment= × =rFsin(θ)=F(rsin(θ))=Fr⊥)
∴FA=2mg/3
Force of magnitude mg is acting downwards at its COM.
Moment balance around B gives:
mg(2L/3)−FA(L)=0
(Moment= × =rFsin(θ)=F(rsin(θ))=Fr⊥)
∴FA=2mg/3
The motion of a potter’s wheel is an example of
- a)rolling motion
- b)rotatory motion
- c)translatory motion
- d)precessional motion
Correct answer is option 'B'. Can you explain this answer?
The motion of a potter’s wheel is an example of
a)
rolling motion
b)
rotatory motion
c)
translatory motion
d)
precessional motion
|
Anjali Iyer answered |
Potter’s wheel is an example of rotary motion. Rotary motion is that kind of motion in which body of the mass moves along a circular path about an axis which remains fixed.
The centre of mass of a system of particles does not depend on
- a)masses of the particles
- b)forces on the particles
- c)relative distances between the particles
- d)position of the particles
Correct answer is option 'B'. Can you explain this answer?
The centre of mass of a system of particles does not depend on
a)
masses of the particles
b)
forces on the particles
c)
relative distances between the particles
d)
position of the particles
|
Neha Joshi answered |
The resultant of all forces, on any system of particles, is zero. Therefore, their centre of mass does not depend upon the forces acting on the particles.
If a shell at rest explodes then the centre of mass of the fragments
- a)remains at rest
- b)Moves along a parabolic path
- c)Moves along a straight line
- d)moves along an elliptical path
Correct answer is option 'A'. Can you explain this answer?
If a shell at rest explodes then the centre of mass of the fragments
a)
remains at rest
b)
Moves along a parabolic path
c)
Moves along a straight line
d)
moves along an elliptical path
|
Gaurav Kumar answered |
As a shell explosion is not governed by any external force, we get that no external force acts upon the shell, and hence it remains at rest.
Two rings have their moment of inertia in the ratio 2:1 and their diameters are in the ratio 2:1. The ratio of their masses will be:- a)1:2
- b)2:1
- c)1:4
- d)1:1
Correct answer is option 'A'. Can you explain this answer?
Two rings have their moment of inertia in the ratio 2:1 and their diameters are in the ratio 2:1. The ratio of their masses will be:
a)
1:2
b)
2:1
c)
1:4
d)
1:1
|
|
Suresh Reddy answered |
We know that MI of a ring is mr2
Where m is mass of the ring and r is its radius
When we have ratio of I = 2:1
And ratio of r = 2:1
We get ratio of r2 = 4:1
Thus to make this ratio 2:1 , that ratio of masses must be 1:2
Where m is mass of the ring and r is its radius
When we have ratio of I = 2:1
And ratio of r = 2:1
We get ratio of r2 = 4:1
Thus to make this ratio 2:1 , that ratio of masses must be 1:2
If a man of mass M jumps to the ground from height h and his centre of mass moves a distance x in the time taken by him to hit the ground, the average force acting on him is (assuming his retardation to be constant during his impact with the ground)- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
If a man of mass M jumps to the ground from height h and his centre of mass moves a distance x in the time taken by him to hit the ground, the average force acting on him is (assuming his retardation to be constant during his impact with the ground)
a)
b)
c)
d)
|
|
Neha Joshi answered |
As the center of mass moves through a distance x, the average force F acting on the man is calculated from :
Work done = Change in potential energy
⇒ Fx = Mgh
⇒ F =
Work done = Change in potential energy
⇒ Fx = Mgh
⇒ F =
In the pulley system shown, if radii of the bigger and smaller pulley are 2 m and 1 m respectively and the acceleration of block ‘A’ is 5 m/s2 in the downward direction , then the acceleration of block B will be :
- a)0 m / s2
- b)5 m / s2
- c) 10 m / s2
- d) 5 / 2 m / s2
Correct answer is option 'D'. Can you explain this answer?
In the pulley system shown, if radii of the bigger and smaller pulley are 2 m and 1 m respectively and the acceleration of block ‘A’ is 5 m/s2 in the downward direction , then the acceleration of block B will be :
a)
0 m / s2
b)
5 m / s2
c)
10 m / s2
d)
5 / 2 m / s2
|
|
Anjali Sharma answered |
Since both pulley are connected to each other. So, their angular acceleration are equal.
⇒α1=α2⇒a1/ r1=a2/ r2 because α=a/r
⇒a1/1=5/2(Given, a2=5m/s2)
⇒a1=5/2 m/s2
Therefore, D is the correct option.
⇒α1=α2⇒a1/ r1=a2/ r2 because α=a/r
⇒a1/1=5/2(Given, a2=5m/s2)
⇒a1=5/2 m/s2
Therefore, D is the correct option.
There are two circular iron discs A and B having masses in the ratio 1:2 and diameter in the ratio 2:1. The ratio of their moment of inertia is- a)4:1
- b)1:3
- c)2:1
- d)8:1
Correct answer is option 'C'. Can you explain this answer?
There are two circular iron discs A and B having masses in the ratio 1:2 and diameter in the ratio 2:1. The ratio of their moment of inertia is
a)
4:1
b)
1:3
c)
2:1
d)
8:1
|
Preeti Iyer answered |
Given,
Mass of A=1,
Mass of B=2.
diameter if A=2,
diameter if B=1.
radius (r) of A=d/2=2/2=1.
radius (r) of B=d/2=1/2.
we know ,
moment of inertia of disc=MR2/2.
moment of inertia (I)of A/moment of inertia (I)of B=MR2/2/MR2/2.
(I) of A/(I) of B=1×12/2/2×(1/2)2/2.
=1×1/2/2×(1/4)/2.
=1/2/(1/2)/2.
=1/2/1/4.
=4/2.
=2/1.
Mass of A=1,
Mass of B=2.
diameter if A=2,
diameter if B=1.
radius (r) of A=d/2=2/2=1.
radius (r) of B=d/2=1/2.
we know ,
moment of inertia of disc=MR2/2.
moment of inertia (I)of A/moment of inertia (I)of B=MR2/2/MR2/2.
(I) of A/(I) of B=1×12/2/2×(1/2)2/2.
=1×1/2/2×(1/4)/2.
=1/2/(1/2)/2.
=1/2/1/4.
=4/2.
=2/1.
When external forces acting on a body are zero, then its centre of mass- a)remains stationary
- b)moves with uniform velocity
- c)either remains stationary or moves with uniform velocity
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
When external forces acting on a body are zero, then its centre of mass
a)
remains stationary
b)
moves with uniform velocity
c)
either remains stationary or moves with uniform velocity
d)
none of these
|
|
Naina Sharma answered |
When force acting upon the body results zero, the resulting acceleration due to net force applied is also zero, and hence by the law of inertia the motion of the body either at rest or constant velocity wont change.
A homogeneous cubical brick lies motionless on a rough inclined surface. The half of the brick which applies greater pressure on the plane is : 
- a) Left half
- b)Right half
- c)Both applies equal pressure
- d)The answer depend upon coefficient of friction
Correct answer is option 'A'. Can you explain this answer?
A homogeneous cubical brick lies motionless on a rough inclined surface. The half of the brick which applies greater pressure on the plane is :
a)
Left half
b)
Right half
c)
Both applies equal pressure
d)
The answer depend upon coefficient of friction
|
Mohit Rajpoot answered |
As weight of the body (mg) acts along the left side i.e. mg sinx acts along flow of object.
mgcosx acts perpendicular to flow of the object.
The pressure of right half comes on left half hence left half has maximum pressure.
mgcosx acts perpendicular to flow of the object.
The pressure of right half comes on left half hence left half has maximum pressure.
There are some passengers inside a stationary railway compartment. The centre of masses of the compartment itself(without the passengers) is C1, while the centre of mass of the compartment plus passengers’ system is C2. if the passengers moves about inside the compartment
- a)both C1 and C2 will move with respect to the ground
- b)neither C1 nor C2 will move with respect to the ground
- c)C1 will move but C2 will be stationary with respect to the ground
- d)C2 will move but C1 will be stationary with respect to the ground
Correct answer is option 'C'. Can you explain this answer?
There are some passengers inside a stationary railway compartment. The centre of masses of the compartment itself(without the passengers) is C1, while the centre of mass of the compartment plus passengers’ system is C2. if the passengers moves about inside the compartment
a)
both C1 and C2 will move with respect to the ground
b)
neither C1 nor C2 will move with respect to the ground
c)
C1 will move but C2 will be stationary with respect to the ground
d)
C2 will move but C1 will be stationary with respect to the ground
|
Lavanya Menon answered |
When net Fexternal=0, then the centre of mass of the system remains at rest.
Thus if the passenger move inside the compartment which donot require any external force, so the centre of mass of the "passenger + compartment" system must remain at rest and hence C2 will be fixed w.r.t ground.
Also due to the movement of the passenger, the position of centre of mass of the passengers only will change, thus C1 will have to move in such a way that C2 may remain fixed w.r.t ground.
Can you explain the answer of this question below:The angular position of a particle (in radians), along a circle of radius 0.8 m is given by the function in time (seconds) by 
. The linear velocity of the particle
- A:
1.84 m/s
- B:
1.80 m/s
- C:
1.82 m/s
- D:
1.88 m/s
The answer is a.
The angular position of a particle (in radians), along a circle of radius 0.8 m is given by the function in time (seconds) by . The linear velocity of the particle
1.84 m/s
1.80 m/s
1.82 m/s
1.88 m/s
|
Knowledge Hub answered |
The answer is option A
So, as we know,
After differentiating angular velocity with respect to t,
Linear velocity=2t+2.3=ω
Now,
Velocity=r x ω(where r is 0.8m)
=0.8x2.3
=1.84m/s
So, as we know,
After differentiating angular velocity with respect to t,
Linear velocity=2t+2.3=ω
Now,
Velocity=r x ω(where r is 0.8m)
=0.8x2.3
=1.84m/s
A solid sphere and a hollow sphere of the same mass have the same moments of inertia about their respective diameters, the ratio of their radii is- a)(5)1/2 : (3)1/2
- b) (3)1/2 : (5)1/2
- c)3 : 2
- d)2 : 3
Correct answer is option 'A'. Can you explain this answer?
A solid sphere and a hollow sphere of the same mass have the same moments of inertia about their respective diameters, the ratio of their radii is
a)
(5)1/2 : (3)1/2
b)
(3)1/2 : (5)1/2
c)
3 : 2
d)
2 : 3
|
|
Gaurav Kumar answered |
We know moment of inertia of solid sphere Is=2/5msRs2 and
moment of inertia of hollow sphere IH=2/3mHRH2 As per question Is=IH
Now,
2/5msRs2=2/3mHRH2
as the masses are equal the ratio of their radii will be
Rs2 /RH2 =2/3/2/5=√5/3=(5)1/2: (3)1/2
moment of inertia of hollow sphere IH=2/3mHRH2 As per question Is=IH
Now,
2/5msRs2=2/3mHRH2
as the masses are equal the ratio of their radii will be
Rs2 /RH2 =2/3/2/5=√5/3=(5)1/2: (3)1/2
A disc of radius b and mass m rolls down an inclined plane of vertical height h. the translational speed when it reaches the bottom of the plane will be - a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
A disc of radius b and mass m rolls down an inclined plane of vertical height h. the translational speed when it reaches the bottom of the plane will be
a)
b)
c)
d)
|
|
Preeti Iyer answered |
The difference in the potential of the body when it rolls down through a vertical height h, is mgh.
As the KE at the top point is zero and let say KE at bottom is ½ mv2 + ½ Iw2
Where m is its mass, I is its moment of inertia, I = ½ mr2
Where r is its radius, v is its gained translational speed and w is its gained angular speed.
w = v/r
Hence equating PE and KE gives
mgh = ½ mv2 + ½ Iw2
That is mgh = ½ mv2 + ½ mr2.(v/r)2
We get mgh = ½ mv2 + ¼ mv2
Thus we get v = √gh/3
As the KE at the top point is zero and let say KE at bottom is ½ mv2 + ½ Iw2
Where m is its mass, I is its moment of inertia, I = ½ mr2
Where r is its radius, v is its gained translational speed and w is its gained angular speed.
w = v/r
Hence equating PE and KE gives
mgh = ½ mv2 + ½ Iw2
That is mgh = ½ mv2 + ½ mr2.(v/r)2
We get mgh = ½ mv2 + ¼ mv2
Thus we get v = √gh/3
A uniform rod of length L and mass M is bent to make a reactangle PQRS such that QR/PQ = 1/2 Moment of inertia of the rectangle about the side QR can be expressed as :
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
A uniform rod of length L and mass M is bent to make a reactangle PQRS such that QR/PQ = 1/2 Moment of inertia of the rectangle about the side QR can be expressed as :
a)
b)
c)
d)
|
|
Sushil Kumar answered |
Firstly, using the relations given in the questions, we will find the length and mass of the parts
A rigid body is one
- a)the sum of distances of all particles from the axis remains constant
- b)in which the distance between all pairs of particles remains fixed
- c)whose centre of mass follows a parabolic path
- d)that deforms and comes back to its original shape after getting deformed
Correct answer is option 'B'. Can you explain this answer?
A rigid body is one
a)
the sum of distances of all particles from the axis remains constant
b)
in which the distance between all pairs of particles remains fixed
c)
whose centre of mass follows a parabolic path
d)
that deforms and comes back to its original shape after getting deformed
|
|
Krishna Iyer answered |
A body is said to be a rigid body if the body remains in its original shape even under the influence of external force. We can also say that if distance between two points of the body does not change with time regardless of external forces exerted on it, then the body is said to be a rigid body.
Calculate the M.I. of a thin uniform ring about an axis tangent to the ring and in a plane of the ring, if its M.I. about an axis passing through the centre and perpendicular to plane is 4 kg m2.- a)12 kg m2
- b)3 kg m2
- c)6 kg m2
- d)9 kg m2
Correct answer is option 'C'. Can you explain this answer?
Calculate the M.I. of a thin uniform ring about an axis tangent to the ring and in a plane of the ring, if its M.I. about an axis passing through the centre and perpendicular to plane is 4 kg m2.
a)
12 kg m2
b)
3 kg m2
c)
6 kg m2
d)
9 kg m2
|
|
Naina Sharma answered |
At tangent I =MR2/2 + MR2 --- (1)
MR2/2 = 4, MR2 = 8 then substitute in (1) you will get I = 12, this is for tangent perpendicular to plane then divide by 2 you will get tangent along the plane
MR2/2 = 4, MR2 = 8 then substitute in (1) you will get I = 12, this is for tangent perpendicular to plane then divide by 2 you will get tangent along the plane
The M.I. of a disc about its diameter is 2 units. Its M.I. about axis through a point on its rim and in the plane of the disc is- a)4 unit
- b)6 unit
- c)8 unit
- d)10 unit
Correct answer is option 'D'. Can you explain this answer?
The M.I. of a disc about its diameter is 2 units. Its M.I. about axis through a point on its rim and in the plane of the disc is
a)
4 unit
b)
6 unit
c)
8 unit
d)
10 unit
|
|
Krishna Iyer answered |
We know that for a disc of mass m and radius r
MI of a disc about its diameter = mr2/4 = 2
And also MI about a point on its rim = mr2/4 + mr2
= 5mr2/4
= 5 x 2 = 10
MI of a disc about its diameter = mr2/4 = 2
And also MI about a point on its rim = mr2/4 + mr2
= 5mr2/4
= 5 x 2 = 10
A CD accelerates uniformly from rest to 200 spins per minute in 8 seconds. If the rate of change of speed is constant, determine the instantaneous acceleration in rad/s2 .- a)200.2 π rad/s2
- b)25 rad/ s2
- c)20.9 rad/ s2
- d)2.6 rad/ s2
Correct answer is option 'D'. Can you explain this answer?
A CD accelerates uniformly from rest to 200 spins per minute in 8 seconds. If the rate of change of speed is constant, determine the instantaneous acceleration in rad/s2 .
a)
200.2 π rad/s2
b)
25 rad/ s2
c)
20.9 rad/ s2
d)
2.6 rad/ s2
|
Varun Kapoor answered |
The correct option is A
2.6 rad/ s2
The moment of inertia and rotational kinetic energy of a fly wheel are 20kg-m2 and 1000 joule respectively. Its angular frequency per minute would be -- a)600/π
- b)25/π2
- c)5/π
- d)300/π
Correct answer is option 'D'. Can you explain this answer?
The moment of inertia and rotational kinetic energy of a fly wheel are 20kg-m2 and 1000 joule respectively. Its angular frequency per minute would be -
a)
600/π
b)
25/π2
c)
5/π
d)
300/π
|
|
Riya Banerjee answered |
Rotational K.E. = ½ × moment of inertia × square of angular velocity = ½Iw2
Four point masses, each of value m, are placed at the corners of a square ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD is
- a) 3 ml2
- b) ml2
- c) 2 ml2
- d) √3 ml2
Correct answer is option 'A'. Can you explain this answer?
Four point masses, each of value m, are placed at the corners of a square ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD is
a)
3 ml2b)
ml2c)
2 ml2d)
√3 ml2|
|
Raghav Bansal answered |
IA=0 (on the Axis .)
IB=ID=mr2
=m(l/√2)2
=ml2/2
IC=m(√2l)2
IC=2ml2
I=IA+IB+IC+ID
I=0+ml2/2+ml2/2+2ml2
I=3ml2
An automobile engine develops 100H.P. when rotating at a speed of 1800 rad/min. The torque it delivers is- a)3.33 W-s
- b)200W-s
- c)248.7 W-s
- d)2487 W-s
Correct answer is option 'D'. Can you explain this answer?
An automobile engine develops 100H.P. when rotating at a speed of 1800 rad/min. The torque it delivers is
a)
3.33 W-s
b)
200W-s
c)
248.7 W-s
d)
2487 W-s
|
Manish Aggarwal answered |
100 HP = 74570 W or 74.57 KW Now, P = 2*π*N*T/60 where, P is the power (in W), N is the operating speed of the engine (in r.p.m.) and T is the Torque (in N.m). Therefore, 74570 = 2*π*1800*T/60 i.e. T = 395.606 N.m
Two rings of radius R and nR made up of same material have the ratio of moment of inertia about an axis passing through centre is 1 : 8. The value of n is- a)2
- b)2√2
- c)4
- d)1/2
Correct answer is option 'A'. Can you explain this answer?
Two rings of radius R and nR made up of same material have the ratio of moment of inertia about an axis passing through centre is 1 : 8. The value of n is
a)
2
b)
2√2
c)
4
d)
1/2
|
Naman Flora answered |
Answer should be B as moment of inertia of ring Mr^2 and Mn^2R^2 Mn^2r^2/Mr^2=1/8 ; solving we get n^2=8 and n=2root2
If three balls of same radius are placed touching each other on a horizontal surface such that they will form an equilateral triangle, when their centers are joined. What will be the position of the centre of mass of the system?- a)At the point of intersection of the medians.
- b)At the line joining the centers of any two balls.
- c)At the centre of one of the ball.
- d)At the horizontal surface.
Correct answer is option 'A'. Can you explain this answer?
If three balls of same radius are placed touching each other on a horizontal surface such that they will form an equilateral triangle, when their centers are joined. What will be the position of the centre of mass of the system?
a)
At the point of intersection of the medians.
b)
At the line joining the centers of any two balls.
c)
At the centre of one of the ball.
d)
At the horizontal surface.
|
|
Neha Joshi answered |
The COM will be at the intersection point of the medians as that is the most symmetric point of this symmetric system.
When a shell was following a parabolic path in the air, it explodes somewhere in its flight. The centre of mass of fragments will continue to move in- a)any direction
- b)horizontal direction
- c)same parabolic path
- d)vertical direction
Correct answer is option 'C'. Can you explain this answer?
When a shell was following a parabolic path in the air, it explodes somewhere in its flight. The centre of mass of fragments will continue to move in
a)
any direction
b)
horizontal direction
c)
same parabolic path
d)
vertical direction
|
|
Riya Banerjee answered |
The internal forces have no effect on the trajectory of the center of mass, and the forces due to explosion are the internal forces. So the center of mass will follow the same parabolic path even after the explosion.
A mixer grinder rotates clockwise, its angular velocity will be :- a)zero
- b)negative
- c)uniform but not zero
- d)positive
Correct answer is option 'B'. Can you explain this answer?
A mixer grinder rotates clockwise, its angular velocity will be :
a)
zero
b)
negative
c)
uniform but not zero
d)
positive
|
Ashish Roy answered |
**Explanation:**
A mixer grinder is a device that is used for grinding and mixing various ingredients. It consists of a motor and a set of blades that rotate at high speeds to perform the grinding and mixing tasks. When the mixer grinder is turned on, the motor starts rotating the blades in a clockwise direction.
**Angular Velocity:**
Angular velocity is a measure of how quickly an object rotates or moves around a central point. It is defined as the rate of change of angular displacement with respect to time. The direction of the angular velocity is determined by the direction of rotation. In the case of a mixer grinder rotating clockwise, the angular velocity will be negative.
**Direction of Angular Velocity:**
The direction of angular velocity is determined by the right-hand rule. According to the right-hand rule, if the fingers of the right hand curl in the direction of rotation, the thumb will point in the direction of the angular velocity vector. In the case of a mixer grinder rotating clockwise, the fingers of the right hand curl in the clockwise direction, and the thumb points in the opposite direction, which is counterclockwise or negative.
**Significance of Negative Angular Velocity:**
A negative angular velocity indicates that the object is rotating in the opposite direction compared to the conventional positive direction. In the case of a mixer grinder, a negative angular velocity means that the blades are rotating counterclockwise when viewed from above. This counterclockwise rotation is necessary for the blades to effectively grind and mix the ingredients.
**Conclusion:**
In conclusion, a mixer grinder rotates clockwise, which means its angular velocity will be negative. The negative angular velocity indicates that the blades are rotating counterclockwise when viewed from above, allowing them to efficiently perform the grinding and mixing tasks.
A mixer grinder is a device that is used for grinding and mixing various ingredients. It consists of a motor and a set of blades that rotate at high speeds to perform the grinding and mixing tasks. When the mixer grinder is turned on, the motor starts rotating the blades in a clockwise direction.
**Angular Velocity:**
Angular velocity is a measure of how quickly an object rotates or moves around a central point. It is defined as the rate of change of angular displacement with respect to time. The direction of the angular velocity is determined by the direction of rotation. In the case of a mixer grinder rotating clockwise, the angular velocity will be negative.
**Direction of Angular Velocity:**
The direction of angular velocity is determined by the right-hand rule. According to the right-hand rule, if the fingers of the right hand curl in the direction of rotation, the thumb will point in the direction of the angular velocity vector. In the case of a mixer grinder rotating clockwise, the fingers of the right hand curl in the clockwise direction, and the thumb points in the opposite direction, which is counterclockwise or negative.
**Significance of Negative Angular Velocity:**
A negative angular velocity indicates that the object is rotating in the opposite direction compared to the conventional positive direction. In the case of a mixer grinder, a negative angular velocity means that the blades are rotating counterclockwise when viewed from above. This counterclockwise rotation is necessary for the blades to effectively grind and mix the ingredients.
**Conclusion:**
In conclusion, a mixer grinder rotates clockwise, which means its angular velocity will be negative. The negative angular velocity indicates that the blades are rotating counterclockwise when viewed from above, allowing them to efficiently perform the grinding and mixing tasks.
A thin rod of length 4 l , mass 4 m is bent at the points as shown in the fig. What is the moment of inertia of the rod about the axis passing through point O and perpendicular to the plane of the paper : 
- a)ml2/3
- b)10ml2/3
- c)ml2/12
- d)ml2/24
Correct answer is option 'B'. Can you explain this answer?
A thin rod of length 4 l , mass 4 m is bent at the points as shown in the fig. What is the moment of inertia of the rod about the axis passing through point O and perpendicular to the plane of the paper :
a)
ml2/3
b)
10ml2/3
c)
ml2/12
d)
ml2/24
|
|
Krishna Iyer answered |
Since the mass of rod is 54m and length is 4l.
so mass C = m
and length of AB = BO=OC= CD = l
WE, know moment of Inertia of a rod about to end = ml2/3
So, moment of Inertia of AB, BO, OC, CD about B, O, O, C respectively = ml2/3
From parallel axis theorem
Moment of Inertia of AB about O.
=ml2/3+ml2=4ml2/3
Similarly od CD about O =4ml2/3
SO moment of Inertia Rod about O
=ml2/3+ml2/3+4ml2/3+4ml2/3
=10ml2/3
so mass C = m
and length of AB = BO=OC= CD = l
WE, know moment of Inertia of a rod about to end = ml2/3
So, moment of Inertia of AB, BO, OC, CD about B, O, O, C respectively = ml2/3
From parallel axis theorem
Moment of Inertia of AB about O.
=ml2/3+ml2=4ml2/3
Similarly od CD about O =4ml2/3
SO moment of Inertia Rod about O
=ml2/3+ml2/3+4ml2/3+4ml2/3
=10ml2/3
Fig shows a flywheel of radius 10 cm. Its moment of inertia about the rotation axis is 0.4 kg m2. A massless string passes over the flywheel and a mass 2 kg is attached at its lower end. Angular acceleration of the mass in rad/s2 is nearly. 
- a)4.8 rad/s2
- b)6.2 rad/s2
- c) 3.2 rad/s2
- d) 9.6 rad/s2
Correct answer is option 'A'. Can you explain this answer?
Fig shows a flywheel of radius 10 cm. Its moment of inertia about the rotation axis is 0.4 kg m2. A massless string passes over the flywheel and a mass 2 kg is attached at its lower end. Angular acceleration of the mass in rad/s2 is nearly.
a)
4.8 rad/s2
b)
6.2 rad/s2
c)
3.2 rad/s2
d)
9.6 rad/s2
|
|
Mohit Rajpoot answered |
I took g = 10 but if g = 9.8 is taken then we get the exact answer.
Can you explain the answer of this question below:An isolated particle of mass m is moving in a horizontal plane (x,y) along the x axis at a certain height above the ground. It suddenly explodes into two fragments of masses m/4 and 3m/4. An instant later, the smaller fragments is at y = +15 cm. The larger fragment at this instant is at
- A:
y = +5 cm
- B:
y = +20 cm
- C:
y = -20 cm
- D:
y = -5 cm
The answer is d.
An isolated particle of mass m is moving in a horizontal plane (x,y) along the x axis at a certain height above the ground. It suddenly explodes into two fragments of masses m/4 and 3m/4. An instant later, the smaller fragments is at y = +15 cm. The larger fragment at this instant is at
y = +5 cm
y = +20 cm
y = -20 cm
y = -5 cm
|
Ambition Institute answered |
As the particle is exploded only due to its internal energy.
⇒ net external force during this process is 0 i.e. center mass will not change.
⇒ net external force during this process is 0 i.e. center mass will not change.
Let the particle while the explosion was above the origin of the coordinate system i.e. just before explosion xcm =0 and ycm =0
After the explosion, the Centre of mass will be at xcm =0 and ycm =0
Since smaller fragment has fallen on the y-axis.
Let positon of larger fragment be y.
Let positon of larger fragment be y.
m * ycm = (m/4 * 15) + (3m/4 * y)
⇒ (m/4 * 15) + (3m/4 * y) = 0
⇒ y = - 5 cm
⇒ (m/4 * 15) + (3m/4 * y) = 0
⇒ y = - 5 cm
An inclined plane makes an angle of 30o with the horizontal. A ring rolling down this inclinedplane from rest without slipping has a linear acceleration equal to :- a)2g/3
- b)g/2
- c)g/3
- d)g/4
Correct answer is option 'D'. Can you explain this answer?
An inclined plane makes an angle of 30o with the horizontal. A ring rolling down this inclinedplane from rest without slipping has a linear acceleration equal to :
a)
2g/3
b)
g/2
c)
g/3
d)
g/4
|
|
Neha Joshi answered |
We know for a rolling body acceleration down an inclined plane, a = g.sinθ / (1 + I/mR2)
Where I is body's moment of inertia. Here, I = mR2
Thus just by putting the values to the formula we get
a = g.sin 30° / 2
= g/4
Where I is body's moment of inertia. Here, I = mR2
Thus just by putting the values to the formula we get
a = g.sin 30° / 2
= g/4
An equilateral prism of mass m rests on a rough horizontal surface with coefficient of friction m . A horizontal force F is applied on the prism as shown in the fig. If the coefficient of friction is sufficiently high so that the prism does not slide before toppling, then the minimum force required to topple the prism is :
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
An equilateral prism of mass m rests on a rough horizontal surface with coefficient of friction m . A horizontal force F is applied on the prism as shown in the fig. If the coefficient of friction is sufficiently high so that the prism does not slide before toppling, then the minimum force required to topple the prism is :
a)
b)
c)
d)
|
|
Rajesh Gupta answered |
As the force is applied towards right the prism will topple at O. Now, for the prism to topple the torque due to force "F" must be greater than or equal to the torque due to weight "Mg" about the point O.
If the net force acting on the system of particles is zero, then which of the following may vary
- a)moment of inertia
- b)Kinetic energy of the system
- c)velocity of centre of mass
- d)position of centre of mass
Correct answer is option 'D'. Can you explain this answer?
If the net force acting on the system of particles is zero, then which of the following may vary
a)
moment of inertia
b)
Kinetic energy of the system
c)
velocity of centre of mass
d)
position of centre of mass
|
|
Suresh Kumar answered |
Correct answer is B . because when a=0 then v=constant hence K.E. will be constant and position of com will vary
A child sits stationary at one end of long trolley moving uniformly with speed v on a smooth horizontal floor. If the child gets up and runs about on the trolley in the forward direction with speed u. The centre of mass of the system (child + trolley) will move with speed- a)v
- b)zero
- c)u + v
- d)v/u
Correct answer is option 'A'. Can you explain this answer?
A child sits stationary at one end of long trolley moving uniformly with speed v on a smooth horizontal floor. If the child gets up and runs about on the trolley in the forward direction with speed u. The centre of mass of the system (child + trolley) will move with speed
a)
v
b)
zero
c)
u + v
d)
v/u
|
|
Preeti Iyer answered |
The child is running arbitrarily on a trolley moving with velocity v. However, the running of the child will produce no effect on the velocity of the centre of mass of the trolley. This is because the force due to the boy’s motion is purely internal. Internal forces produce no effect on the motion of the bodies on which they act. Since no external force is involved in the boy–trolley system, the boy’s motion will produce no change in the velocity of the centre of mass of the trolley.
Three rings, each of mass P and radius Q are arranged as shown in the figure. The moment of inertia of the arrangement about YY' axis will be 
- a)
PQ2 - b)
PQ2 - c)
PQ2 - d)
PQ2
Correct answer is option 'A'. Can you explain this answer?
Three rings, each of mass P and radius Q are arranged as shown in the figure. The moment of inertia of the arrangement about YY' axis will be
a)
PQ2b)
PQ2c)
PQ2d)
PQ2
|
Lohit Matani answered |
For ring 1
(MOI)1 = MOI about diameter + PQ2
MOI about diameter = ½ PQ2
(MOI)1 = 3/2 PQ2
Similarly, (MOI)2 = 3/2 PQ2
(MOI)3 = ½ PQ2
Total MOI = 7/2 PQ2
(MOI)1 = MOI about diameter + PQ2
MOI about diameter = ½ PQ2
(MOI)1 = 3/2 PQ2
Similarly, (MOI)2 = 3/2 PQ2
(MOI)3 = ½ PQ2
Total MOI = 7/2 PQ2
A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :- a)Downwards at any point between A and B.
- b) Downwards at mid point of AB.
- c)Downwards at a point C such that AC = 1m.
- d) Downwards at a point D such that BD = 1m.
Correct answer is option 'D'. Can you explain this answer?
A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :
a)
Downwards at any point between A and B.
b)
Downwards at mid point of AB.
c)
Downwards at a point C such that AC = 1m.
d)
Downwards at a point D such that BD = 1m.
|
|
Riya Banerjee answered |
The moment of inertia of a body depends upon -- a) mass only
- b) angular velocity only
- c)distribution of particles only
- d)mass and distribution of mass about the axis
Correct answer is option 'D'. Can you explain this answer?
The moment of inertia of a body depends upon -
a)
mass only
b)
angular velocity only
c)
distribution of particles only
d)
mass and distribution of mass about the axis
|
|
Preeti Iyer answered |
I = mr2, therefore it depends on body mass and its mass distribution.
An object comprises of a uniform ring of radius R and its uniform chord AB (not necessarily made of the same material) as shown. Which of the following can not be the centre of mass of the object.
- a) (R/3, R/3)
- b) (R/3, R/2)
- c)(R/4, R/4)
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
An object comprises of a uniform ring of radius R and its uniform chord AB (not necessarily made of the same material) as shown. Which of the following can not be the centre of mass of the object.
a)
(R/3, R/3)
b)
(R/3, R/2)
c)
(R/4, R/4)
d)
None of these
|
|
Raghav Bansal answered |
Since the points must have symmetric coordinates so B is the answer.
A wheel starting with angular velocity of 10 radian/sec acquires angular velocity of 100 radian/sec in 15 seconds. If moment of inertia is 10kg-m2, then applied torque (in newton-metre) is- a)900
- b)100
- c)90
- d)60
Correct answer is option 'D'. Can you explain this answer?
A wheel starting with angular velocity of 10 radian/sec acquires angular velocity of 100 radian/sec in 15 seconds. If moment of inertia is 10kg-m2, then applied torque (in newton-metre) is
a)
900
b)
100
c)
90
d)
60
|
|
Gaurav Kumar answered |
I(ωf - ωi)/t = τ
Hence, τ = 60
Hence, τ = 60
Consider the following statementsAssertion (A) : The moment of inertia of a rigid body reduces to its minimum value as compared to any other parallel axis when the axis of rotation passes through its centre of mass.Reason (R) : The weight of a rigid body always acts through its centre of mass in uniform gravitational field. Of these statements :- a)both A and R are true and R is the correct explanation of A
- b) both A and R are true but R is not a correct explanation of A
- c)A is true but R is false
- d) A is false but R is true
Correct answer is option 'B'. Can you explain this answer?
Consider the following statements
Assertion (A) : The moment of inertia of a rigid body reduces to its minimum value as compared to any other parallel axis when the axis of rotation passes through its centre of mass.
Reason (R) : The weight of a rigid body always acts through its centre of mass in uniform gravitational field. Of these statements :
a)
both A and R are true and R is the correct explanation of A
b)
both A and R are true but R is not a correct explanation of A
c)
A is true but R is false
d)
A is false but R is true
|
|
Neha Joshi answered |
Moment of inertia about its centroidal axis has a minimum value as the centroidal axis has mass evenly distributed around it thereby providing minimum resistance to rotation as compared to any other axis. Also the statement-2 is correct but is not the correct explanation for statement-1.
On applying a constant torque on a body- a) Linear velocity may be increases
- b)Angular velocity may be increases
- c)It will rotate with constant angular velocity
- d)It will move with constant velocity
Correct answer is option 'A'. Can you explain this answer?
On applying a constant torque on a body
a)
Linear velocity may be increases
b)
Angular velocity may be increases
c)
It will rotate with constant angular velocity
d)
It will move with constant velocity
|
|
Geetika Shah answered |
If a constant torque is applied it is possible that a positive angular acceleration gets generated which can generate a positive acceleration and hence increasing both velocity and angular velocity.
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