1 Crore+ students have signed up on EduRev. Have you? 
Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2 I respectively about the common axis. Disc A is imparted an initial angular velocity 2 ω using the entire potential energy of a spring compressed by a distance x_{l}. Disc B is imparted an angular velocity ω by a spring having the same spring constant and compressed by a distance x_{2} .Both the discs rotate in the clockwise direction.
Q1.The ratio x_{1}/x_{2} is
For disc A
For disc B
On dividing (i) and (ii), we get
Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2 I respectively about the common axis. Disc A is imparted an initial angular velocity 2 ω using the entire potential energy of a spring compressed by a distance x_{l}. Disc B is imparted an angular velocity ω by a spring having the same spring constant and compressed by a distance x_{2} .Both the discs rotate in the clockwise direction.
Q.2. When disc B is brought in contact with disc A, they acquire a common angular velocity in time t . The average frictional torque on one disc by the other during this period is
When disc B is brought in contact with disc A Let ω' be the final angular velocity of both the disc rotating together. Apply conservation of angular momentum for the two disc system.
Torque on disc A
Note : The negative sign represents that the torque creates angular retardation.
Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2 I respectively about the common axis. Disc A is imparted an initial angular velocity 2 ω using the entire potential energy of a spring compressed by a distance x_{l}. Disc B is imparted an angular velocity ω by a spring having the same spring constant and compressed by a distance x_{2} .Both the discs rotate in the clockwise direction.
Q.3. The loss of kinetic energy in the above process is
Loss in kinetic energy = (K.E.)_{initial} – (K.E.)_{final}
A uniform thin cylindrical disk of mass M and radius R is attached to two identical massless springs of spring constant k which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance d from its centre. The axle is massless and both the springs and the axle are in horizontal plane.
The unstretched length of each spring is L. The disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping T he coefficient of friction is µ.
Q.4. The net external force acting on the disk when its centre of mass is at displacement x with respect to its equilibrium position is
Wh en th e disc is at a distance x from the mean position (equilibrium position), the forces acting on the disc are shown in the figure
∴ –2kx + f = –Mac ...(1) where ac = acceleration of center
of mass. Also the torque acting on the disc about its center of mass C is
From (i) & (ii)
⇒ Net external force acting on the disc when its centre of mass is at displacement x with respect to the equilibrium position = 4kx/3
directed towards the equilibrium.
A uniform thin cylindrical disk of mass M and radius R is attached to two identical massless springs of spring constant k which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance d from its centre. The axle is massless and both the springs and the axle are in horizontal plane.
The unstretched length of each spring is L. The disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping T he coefficient of friction is µ.
Q.5.The centre of mass of the disk undergoes simple harmonic motion with angular frequency w equal to –
As derived in ans 4.
A uniform thin cylindrical disk of mass M and radius R is attached to two identical massless springs of spring constant k which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance d from its centre. The axle is massless and both the springs and the axle are in horizontal plane.
The unstretched length of each spring is L. The disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping T he coefficient of friction is µ.
The maximum value of V0 for which the disk will roll without slipping is –
From (i) & (ii)
We see that the frictional force depends on x. As x increases, f increases. Also, the frictional force is maximum at x = A where A is the amplitude of S.H.M.
Therefore the maximum frictional force
The force should be utmost equal to the limiting friction (μMg) for rolling without slipping.
The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass.
These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless, stick, as shown in the figure. When the discstick system is rotated about the origin on a horizontal frictionless plane with angular speedω, the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the zaxis and (ii) a rotation of the disc through an instantaneous vertical axis passing through its centre of mass (as is seen from the changed orientation of points P and Q). Both these motions have the same angular speed w in this case
Now consider two similar systems as shown in the figure: Case (a) the disc with its face vertical and parallel to xz plane; Case (b) the disc with its face making an angle of 45° with xy plane and its horizontal diameter parallel to xaxis. In both the cases, the disc is welded at point P, and the systems are rotated with constant angular speed ω about the zaxis.
Q.7. Which of the following statements about the instantaneous axis (passing through the centre of mass) is correct?
Axis of rotation is parallel to zaxis.
The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass.
These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless, stick, as shown in the figure. When the discstick system is rotated about the origin on a horizontal frictionless plane with angular speedω, the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the zaxis and (ii) a rotation of the disc through an instantaneous vertical axis passing through its centre of mass (as is seen from the changed orientation of points P and Q). Both these motions have the same angular speed w in this case
Now consider two similar systems as shown in the figure: Case (a) the disc with its face vertical and parallel to xz plane; Case (b) the disc with its face making an angle of 45° with xy plane and its horizontal diameter parallel to xaxis. In both the cases, the disc is welded at point P, and the systems are rotated with constant angular speed ω about the zaxis.
Q.8.Which of the following statements regarding the angular speed about the instantaneous axis (passing through the centre of mass) is correct?
Since the body is rigid, w is same for any point of the body.
STATEMENT1: If there is no external torque on a body about its center of mass, then the velocity of the center of mass remains constant.
STATEMENT2: The linear momentum of an isolated system remains constant.
Statement 1 : For velocity of centre of mass to remain constant the net force acting on a body must be zero.
Therefore the statement 1 is false.
Statement 2 : The linear momentum of an isolated system remains constant. This statement is true.
STATEMENT1 : Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first.
STATEMENT2 : By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.
The acceleration of a body rolling down an inclined plane is given by
⇒ Acceleration of solid cylinder is more than hollow cylinder and therefore solid cylinder will reach the bottom of the inclined plane first.
∴ Statement 1 is false
• Statement  2 In the case of rolling there will be no heat losses.
Therefore total mechanical energy remains conserved.
The potential energy therefore gets converted into kinetic energy. In both the cases since the initial potential energy is same, the final kinetic energy will also be same. Therefore statement 2 is correct.
Initial angular velocity of a circular disc of mass M is ω 1.
Then two small spheres of mass m are attached gently to diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc?
When two small spheres of mass m are attached gently, the external torque, about the axis of rotation, is zero and therefore the angular momentum about the axis of rotation is constant
The minimum velocity (in ms^{–1}) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is
For negotiating a circular curve on a levelled road, the maximum velocity of the car is v_{max }
A cylinder of height 20 m is completely filled with water. The velocity of efflux of water (in ms^{–1}) through a small hole on the side wall of the cylinder near its bottom is
The velocity of efflux is given
v = √2gh
Where h is the height of the free surface of liquid from the hole
Two identical particles move towards each other with velocity 2v and v respectively. The velocity of centre of mass is
The velocity of centre of mass of two particle system is given by
A solid sphere, a hollow sphere and a ring are released from top of an inclined plane (frictionless) so that they slide down the plane. Then maximum acceleration down the plane is for (no rolling)
This is a case of sliding (the plane being frictionless) and therefore the acceleration of all the bodies is same (g sinθ).
Moment of inertia of a circular wire of mass M and radius R about its diameter is
M. I of a circular wire about an axis nn' passing through the centre of the circle and perpendicular to the plane of the circle = MR^{2}
As shown in the figure, Xaxis and Yaxis lie in the plane of the ring . Then by perpendicular axis theorem
A particle of mass m moves along line PC with velocity v as shown. What is the angular momentum of the particle about P?
Angular momentum (L) = (linear momentum) × (perpendicular distance of the line of action ofmomentum from the axis of rotation) = mv × r [Here r = 0 because the line of = mv × 0 action of momentum passes = 0 through the axis of rotation]
A circular disc X of radius R is made from an iron plate of thickness t, and another disc Y of radius 4R is made from an iron plate of thickness 4
t . Then the relation between the moment of inertia I_{X} and I_{Y} is
We know that density (d)=mass M/volumeV
The moment of inertia of a disc is given by
A particle performing uniform circular motion has angular frequency is doubled & its kinetic energy halved, then the new angular momentum is
be the force acting on a particle having position vector be the torque of this force about the origin.
Then
We know that The angle between tr and rr is 90° and the angle between tr and We also know that the dot product of two vectors which have an angle of 90° between them is zero. Therefore (d) is the correct option
A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same which one of the following will not be affected ?
Angular momentum will remain the same since external torque is zero.
One solid sphere A and another hollow sphere B are of same mass and same outer radii. Their moment of inertia about their diameters are respectively I A and I B Such that
The moment of inertia of solid sphere A about its diameter
The moment of inertia of a hollow sphere B about its diameter
A body A of mass M while falling vertically downwards under gravity breaks into two parts; a body B of mass 1/3 M and a body C of mass 2/3 M. The centre of mass of bodies B and C taken together shifts compared to that of body A towards
Does not shift as no external force acts. The centre of mass of the system continues its original path. It is only the internal forces which comes into play while breaking.
The moment of inertia of a uniform semicircular disc of mass M and radius r about a line perpendicular to the plane of the disc through the centre is
The disc may be assumed as combination of two semi circular parts.
Let I be the moment of inertia of the uniform semicircular disc
A ‘ T’ shaped object with dimensions shown in the figure, is lying on a smooth floor. is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C.
To have linear motion, the has to be applied
at centre of mass. i.e. the point ‘P’has to be at the centre of mass
Consider a two particle system with particles having masses m_{1} and m_{2}. If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle is moved, so as to keep the centre of mass at the same position?
Initially,
Finally, The centre of mass is at the origin
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
I_{nn' }= M.I due to the point mass at B + M.I due to the point mass at D + M.I due to the point mass at C.
= ml^{2 }+ 2ml^{2}=3ml^{2}
A force of ˆ acts on O, the origin of the coordinate system. The torque about the point (1, –1) is
A thin circular ring of mass m and radius R is rotating about its axis with a constant angular velocity ω. Two objects each of mass M are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity ω' =
Applying conservation of angular momentum I' ω' = I ω
A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumferences of the discs coincide. The centre of mass of the new disc is α / R form the centre of the bigger disc. The value of α is
Let the mass per unit area beσ.
Then the mass of the complete disc
The mass of the removed disc Let us consider the above situation to be a complete disc of radius 2R on which a disc of radius R of negative mass is superimposed. Let O be the origin. Then the above figure can be redrawn keeping in mind the concept of centre of mass as :
A round uniform body of radius R, mass M and moment of inertia I rolls down (without slipping) an inclined plane making an angle q with the horizontal. Then its acceleration is
This is a standard formula and should be memorized.
Angular momentum of the particle rotating with a central force is constant due to
We know that
of the body. Central forces act along the center of mass.
Therefore torque about center of mass is zero.
For the given uniform square lamina ABCD, whose centre is O,
By the theorem of perpendicular axes,
Again, by the same theorem
From (i) and (ii), we get I_{EF} = I_{AC}
A thin rod of length ‘L’ is lying along the xaxis with its ends at x = 0 and x = L. Its linear density (mass/length) varies with
, where n can be zero or any positive number. If the position x_{CM} of the centre of mass of the rod is plotted against ‘n’, which of the following graphs best approximates the dependence of x_{CM} on n?
· When n = 0, x = k where k is a constant. This means that the linear mass density is constant. In this case the centre of mass will be at the midelle of the rod ie at L/2. Therefore (c) is ruled out · n is positive and as its value increases, the rate of increase of linear mass density with increase in x increases. This shows that the centre of mass will shift towards that end of the rod where n = L as the value of n increases. Therefore graph (b) is ruled out. · The linear mass density
With increase in the value of n, the centre of mass shift towards the end x = L such that first the shifting is at a higher rate with increase in the value of n and then the rate decreases with the value of n.
These characteristics are represented by graph (a).
Consider a uniform square plate of side ‘a’ and mass ‘m’. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is
According to parallel axis theorem
A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is w. Its centre of mass rises to a maximum height of :
The moment of inertia
of the rod about O is
angular speed of the rod is when the rod is inst ant aneo us ly vertical. The energy of the rod in this condition is where I is the moment of inertia of the rod about O. When the rod is in its extreme portion, its angular velocity is zero momentarily. In this case, the energy of the rod is mgh where h is the maximum height to which the centre of mass (C.M) rises
A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R. Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m, if the string does n ot slip on the pulley, i s:
For translational motion, mg – T = ma .....(1) For rotational motion,
A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc.
As insect moves along a diameter, the effective mass and hence the M.I. first decreases then increases so from principle of conservation of angular momentum, angular speed, first increases then decreases.
A pulley of radius 2 m is rotated about its axis by a force F = (20t – 5t^{2}) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kgm^{2} the number of rotations made by the pulley before its direction of motion is reversed, is:
A hoop of radius r and mass m rotating with an angular velocity ω_{0} is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero.What will be the velocity of the centre of the hoop when it ceases to slip ?
From conservation of angular momentum about any fix point on the surface,
A bob of mass m attached to an inextensible string of length l is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed ω rad/s about the vertical.About the point of suspension:
Torque working on the bob of mass m is , ......(Direction parallel to plane of rotation of particle)
direction of L changes but magnitude remains same.
Distance of the centre of mass of a solid uniform cone from its vertex is z_{0}. If the radius of its base is R and its height is h then z_{0} is equal to :
From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its faces is :
Moment of inertia of the cube about the given axis,
A roller is made by joining together two cones at their vertices O. It is kept on two rails AB and CD, which are placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and Cd (see figure).It is given a light push so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to :
As shown in the diagram, the normal reaction of AB on roller will shift towards O.
This will lead to tending of the system of cones to turn left.
257 videos633 docs256 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
257 videos633 docs256 tests







