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Torque required, τ = 180 Nm
The power of the rotor (P) is related to torque and angular speed by the relation:
P = τω
= 180 × 200 = 36 × 10^{3}
= 36 kW
Hence, the power required by the engine is 36 kW.
Question 7.16. From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
Answer
Mass per unit area of the original disc = σ
Radius of the original disc = R
Mass of the original disc, M = πR^{2}σ
The disc with the cut portion is shown in the following figure:
Mass of the smaller disc, M^{’} = π (R/2)^{2}σ = π R^{2}σ / 4 = M / 4
Let O and O′ be the respective centres of the original disc and the disc cut off from the original. As per the definition of the centre of mass, the centre of mass of the original disc is supposed to be concentrated at O, while that of the smaller disc is supposed to be concentrated at O′.
It is given that:
OO′= R/2
After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses. The two masses are:
M (concentrated at O), and –M′ (=M/4) concentrated at O′
(The negative sign indicates that this portion has been removed from the original disc.)
Let x be the distance through which the centre of mass of the remaining portion shifts from point O.
The relation between the centres of masses of two masses is given as:
x = (m_{1} r_{1} + m_{2} r_{2}) / (m_{1 }+ m_{2})
For the given system, we can write:
x = [ M × 0  M' × (R/2) ] / ( M + (M') ) = R / 6
(The negative sign indicates that the centre of mass gets shifted toward the left of point O.)
Question 7.17. A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?
Answer
Let W and W′ be the respective weights of the metre stick and the coin.
Mass of the meter stick = m^{’}
Mass of each coin, m = 5 g
When the coins are placed 12 cm away from the end P, the centre of mass gets shifted by 5 cm from point R toward the end P. The centre of mass is located at a distance of 45 cm from point P.
The net torque will be conserved for rotational equilibrium about point R.
10 × g(45  12)  m'g(50  45) = 0
∴ m' = 66 g
Hence, the mass of the metre stick is 66 g.
Question 7.18. A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination. (a) Will it reach the bottom with the same speed in each case? (b) Will it take longer to roll down one plane than the other? (c) If so, which one and why?
(a) Mass of the sphere = m
Height of the plane = h
Velocity of the sphere at the bottom of the plane = v
At the top of the plane, the total energy of the sphere = Potential energy = mgh
At the bottom of the plane, the sphere has both translational and rotational kinetic energies.
Hence, total energy = (1/2)mv^{2} + (1/2) I ω^{2}
Using the law of conservation of energy, we can write:
(1/2)mv^{2} + (1/2) I ω^{2} = mgh
For a solid sphere, the moment of inertia about its centre, I = (2/5)mr^{2}
Hence, equation (i) becomes:
(1/2)mr^{2} + (1/2) [(2/5)mr^{2}]ω^{2} = mgh
(1/2)v^{2} + (1/5)r^{2}ω^{2} = gh
But we have the relation, v = rω
∴ (1/2)v^{2} + (1/5)v^{2} = gh
v^{2}(7/10) = gh
v = √(10/7)gh
Hence, the velocity of the sphere at the bottom depends only on height (h) and acceleration due to gravity (g). Both these values are constants. Therefore, the velocity at the bottom remains the same from whichever inclined plane the sphere is rolled.
(b) Consider two inclined planes with inclinations θ_{1 }and θ_{2}, related as:
θ_{1} < θ_{2}
The acceleration produced in the sphere when it rolls down the plane inclined at θ_{1} is:
g sin θ_{1}
The various forces acting on the sphere are shown in the following figure.
Similarly, the acceleration produced in the sphere when it rolls down the plane inclined at θ_{2} is:
g sin θ_{2}
The various forces acting on the sphere are shown in the following figure.
R_{2} is the normal reaction to the sphere.
θ_{2} > θ_{1}; sin θ_{2} > sin θ_{1} ... (i)
∴ a_{2} > a_{1 … }(ii)
Initial velocity, u = 0
Final velocity, v = Constant
Using the first equation of motion, we can obtain the time of roll as:
v = u + at
∴ t ∝ (1/α)
For inclination θ_{1} : t_{1} ∝ (1/α_{1})
For inclination θ_{2} : t_{2} ∝ (1/α_{2})
From above equations, we get:
t_{2 }< t_{1}
Hence, the sphere will take a longer time to reach the bottom of the inclined plane having the smaller inclination.
Question 7.19. A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it?
Mass of the hoop, m = 100 kg
Velocity of the hoop, v = 20 cm/s = 0.2 m/s
Total energy of the hoop = Translational K.E. + Rotational K.E.
E_{T} = (1/2)mv^{2} + (1/2) I ω^{2}
Moment of inertia of the hoop about its centre, I = mr^{2}
E_{T} = (1/2)mv^{2} + (1/2) (mr^{2})ω^{2}
But we have the relation, v = rω
∴ E_{T} = (1/2)mv^{2} + (1/2)mr^{2}ω^{2}
= (1/2)mv^{2} + (1/2)mv^{2 }= mv^{2}
The work required to be done for stopping the hoop is equal to the total energy of the hoop.
∴ Required work to be done, W = mv^{2 }= 100 × (0.2)^{2 }= 4 J.
Question 7.20. The oxygen molecule has a mass of 5.30 × 10^{–26} kg and a moment of inertia of 1.94×10^{–46}kg m^{2} about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.
Moment of inertia, I = 1.94 × 10^{–46} kg m^{2}
Velocity of the oxygen molecule, v = 500 m/s
The separation between the two atoms of the oxygen molecule = 2r
Mass of each oxygen atom = m/2
Hence, moment of inertia I, is calculated as:
(m/2)r^{2} + (m/2)r^{2} = mr^{2}
r = ( I / m)^{1/2}
(1.94 × 10^{46} / 5.36 × 10^{26} )^{1/2} = 0.60 × 10^{10} m
It is given that:
KE_{rot} = (2/3)KE_{trans}
(1/2) I ω^{2} = (2/3) × (1/2) × mv^{2}
mr^{2}ω^{2} = (2/3)mv^{2}
ω = (2/3)^{1/2} (v/r)
= (2/3)^{1/2} (500 / 0.6 × 10^{10}) = 6.80 × 10^{12} rad/s.
Question 7.21. A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.
From, 1/2 mv2 + 1/2 I ω2 = mgh
1/2mv2 + 1/2 (1/2 mr2) ω2 = mgh
3/4 mv2 = mgh
h = 3v2/4g = 3 × 52/4 × 9.8 = 1.913 m
If s is the distance up the inclined plane, then as
sin θ = h/s
s = h/sin θ = 1.913/sin 30° = 3.826 m
Time taken to return to the bottom
(Hint: Consider the equilibrium of each side of the ladder separately.)
Moment of inertia when the man stretches his hands to a distance of 90 cm:
2 × m r^{2}
= 2 × 5 × (0.9)^{2}
= 8.1 kg m^{2}
Initial moment of inertia of the system, I_{i} = 7.6 + 8.1 = 15.7 kg m^{2}
Angular speed, ω_{i} = 300 rev/min
Angular momentum, L_{i} = I_{i}ω_{i} = 15.7 × 30 ....(i)
Moment of inertia when the man folds his hands to a distance of 20 cm:
2 × mr^{2}
= 2 × 5 (0.2)^{2} = 0.4 kg m^{2}
Final moment of inertia, I_{f} = 7.6 + 0.4 = 8 kg m^{2}
Final angular speed = ω_{f}
Final angular momentum, L_{f} = I_{f}ω_{f} = 0.79 ω_{f} ...... (ii)
From the conservation of angular momentum, we have:
I_{i}ω_{i} = I_{f}ω_{f}
∴ ω_{f} = 15.7 × 30 / 8 = 58.88 rev/min
(b) Kinetic energy is not conserved in the given process. In fact, with the decrease in the moment of inertia, kinetic energy increases. The additional kinetic energy comes from the work done by the man to fold his hands toward himself.
Question 7.24. A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.
(Hint: The moment of inertia of the door about the vertical axis at one end is ML^{2}/3.)
Answer
Velocity of the bullet, v = 500 m/s
Thickness of the door, L = 1 m
Radius of the door, r = m / 2
Mass of the door, M = 12 kg
Angular momentum imparted by the bullet on the door:
α = mvr
= (10 × 10^{3} ) × (500) × (1/2) = 2.5 kg m^{2} s^{1} ...(i)
Moment of inertia of the door:
I = ML^{2} / 3
= (1/3) × 12 × 1^{2} = 4 kgm^{2}
But α = Iω
∴ ω = α / I
= 2.5 / 4
= 0.625 rad s^{1}
^{}Question 7.25. Two discs of moments of inertia I_{1} and I_{2} about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds ω_{1} and ω_{2} are brought into contact face to face with their axes of rotation coincident. (a) What is the angular speed of the twodisc system? (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take ω_{1} ≠ ω_{2}.
Angular speed of disc I = ω_{1}
Moment of inertia of disc I = I_{2}
Angular speed of disc I = ω_{2}
Angular momentum of disc I, L_{1} = I_{1}ω_{1}
Angular momentum of disc II, L_{2} = I_{2}ω_{2}
Total initial angular momentum L_{i} = I_{1}ω_{1} + I_{2}ω_{2}
When the two discs are joined together, their moments of inertia get added up.
Moment of inertia of the system of two discs, I = I_{1} + I_{2}
Let ω be the angular speed of the system.
Total final angular momentum, L_{T} = (I_{1} + I_{2}) ω
Using the law of conservation of angular momentum, we have:
L_{i} = L_{T}
I_{1}ω_{1} + I_{2}ω_{2} = (I_{1} + I_{2})ω
∴ ω = (I_{1}ω_{1} + I_{2}ω_{2}) / (I_{1} + I_{2})
(b) Kinetic energy of disc I, E_{1} = (1/2) I_{1}ω_{1}^{2}
Kinetic energy of disc II, E_{1} = (1/2) I_{2}ω_{2}^{2}
Total initial kinetic energy, E_{i} = (1/2) ( I_{1}ω_{1}^{2} + I_{2}ω_{2}^{2})
When the discs are joined, their moments of inertia get added up.
Moment of inertia of the system, I = I_{1} + I_{2}
Angular speed of the system = ω
Final kinetic energy E_{f}: = (1/2) ( I_{1} + I_{2}) ω^{2}
= (1/2) ( I_{1} + I_{2}) [ (I_{1}ω_{1} + I_{2}ω_{2}) / (I_{1} + I_{2}) ]^{2}
= (1/2) (I_{1}ω_{1} + I_{2}ω_{2})^{2} / (I_{1} + I_{2})
∴ E_{i}  E_{f}
= (1/2) ( I_{1}ω_{1}^{2} + I_{2}ω_{2}^{2})  (1/2) (I_{1}ω_{1} + I_{2}ω_{2})^{2} / (I_{1} + I_{2})
Solving the equation, we get
= I_{1} I_{2} (ω_{1}  ω_{2})^{2} / 2(I_{1} + I_{2})
All the quantities on RHS are positive
∴ E_{i}  E_{f} > 0
E_{i} > E_{f}
The loss of KE can be attributed to the frictional force that comes into play when the two discs come in contact with each other.
Question 7.26. (a) Prove the theorem of perpendicular axes.
(Hint: Square of the distance of a point (x, y) in the x–y plane from an axis through the origin perpendicular to the plane is x^{2 }+ y^{2}).
(b) Prove the theorem of parallel axes.
(Hint: If the centre of mass is chosen to be the origin ∑ m_{i}r_{i} = 0).
Answer
(a) The theorem of perpendicular axes states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.
A physical body with centre O and a point mass m,in the x–y plane at (x, y) is shown in the following figure.
Moment of inertia about yaxis, I_{y} = my^{2}
Moment of inertia about zaxis, I_{z} = m(x^{2} + y^{2})^{1/2}
I_{x} + I_{y} = mx^{2} + my^{2}
= m(x^{2} + y^{2})
= m [(x^{2} + y^{2})^{1/2}]^{1/2}
I_{x} + I_{y} = I_{z}
Hence, the theorem is proved.
(b) The theorem of parallel axes states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.
Suppose a rigid body is made up of n particles, having masses m_{1}, m_{2}, m_{3}, … , m_{n}, at perpendicular distances r_{1}, r_{2}, r_{3}, … , r_{n} respectively from the centre of mass O of the rigid body.
The moment of inertia about axis RS passing through the point O:
Question 7.27. Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by v^{2} = 2gh / [1 + (k^{2}/R^{2}) ]
Using dynamical consideration (i.e. by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.
Answer
R = Radius of the body
K = Radius of gyration of the body
v = Translational velocity of the body
h =Height of the inclined plane
g = Acceleration due to gravity
Total energy at the top of the plane, E_{1}= mgh
Total energy at the bottom of the plane, E_{b} = KE_{rot} + KE_{trans}
= (1/2) I ω^{2} + (1/2) mv^{2}
But I = mk^{2} and ω = v / R
∴ E_{b} = (1/2)(mk^{2})(v^{2}/R^{2}) + (1/2)mv^{2}
= (1/2)mv^{2} (1 + k^{2} / R^{2})
From the law of conservation of energy, we have:
E_{T} = E_{b}
mgh = (1/2)mv^{2} (1 + k^{2} / R^{2})
∴ v = 2gh / (1 + k^{2} / R^{2})
Hence, the given result is proved.
Question 9.28. A disc rotating about its axis with angular speed ω_{o}is placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in Fig. 7.41? Will the disc roll in the direction indicated?
The disc will not roll
Angular speed of the disc = ω_{o }
Radius of the disc = R
Using the relation for linear velocity, v = ω_{o}R
For point A:
v_{A} = Rω_{o}; in the direction tangential to the right
For point B:
v_{B} = Rω_{o}; in the direction tangential to the left
For point C:
v_{c} = (R/2)ω_{o} in the direction same as that of v_{A}
The directions of motion of points A, B, and C on the disc are shown in the following figure
Since the disc is placed on a frictionless table, it will not roll. This is because the presence of friction is essential for the rolling of a body.
(b) What is the force of friction after perfect rolling begins?
Initial angular speed, ω_{0 }=10 π rad s^{–1}
Coefficient of kinetic friction, μ_{k} = 0.2
Initial velocity of both the objects, u = 0
Motion of the two objects is caused by frictional force. As per Newton’s second law of motion, we have frictional force, f = ma
μ_{k}mg= ma
Where,
a = Acceleration produced in the objects
m = Mass
∴ a = μ_{k}g … (i)
As per the first equation of motion, the final velocity of the objects can be obtained as:
v = u + at
= 0 + μ_{k}gt
= μ_{k}gt … (ii)
The torque applied by the frictional force will act in perpendicularly outward direction and cause reduction in the initial angular speed.
Torque, τ= –Iα
α = Angular acceleration
μ_{k}mgr = –Iα
∴ α = μ_{k}mgr / I .....(iii)
Using the first equation of rotational motion to obtain the final angular speed:
ω = ω_{0} + αt
= ω_{0} + (μ_{k}mgr / I)t ....(iv)
Rolling starts when linear velocity, v = rω
∴ v = r (ω_{0}  μ_{k}mgrt / I) ...(v)
Equating equations (ii) and (v), we get:
μ_{k}gt = r (ω_{0}  μ_{k}mgrt / I)
= rω_{0}  μ_{k}mgr^{2}t / I ....(vi)
For the ring:
I = mr^{2}
∴ μ_{k}gt = rω_{0}  μ_{k}mgr^{2}t / mr^{2}
= rω_{0}  μ_{k}gt
2μ_{k}gt = rω_{0}
∴ t = rω_{0} / 2μ_{k}g
= 0.1 × 10 × 3.14 / 2 × 0.2 × 9.8 = 0.80 s ....(vii)
For the disc: I = (1/2)mr^{2}
∴ μ_{k}gt = rω_{0}  μ_{k}mgr^{2}t / (1/2)mr^{2}
= rω_{0}  2μ_{k}gt
3μ_{k}gt = rω_{0}
∴ t = rω_{0} / 3μ_{k}g
= 0.1 × 10 × 3.14 / 3 × 0.2 × 9.8 = 0.53 s .....(viii)
Since t_{d} > t_{r}, the disc will start rolling before the ring.
Question 7.31. A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction µ_{s} = 0.25.
(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against friction during rolling?
(c) If the inclination θ_{ }of the plane is increased, at what value of θ does the cylinder begin to skid, and not roll perfectly?
Answer
Radius of the cylinder, r = 15 cm = 0.15 m
Coefficient of kinetic friction, µ_{k }= 0.25
Angle of inclination, θ = 30°
Moment of inertia of a solid cylinder about its geometric axis, I = (1/2)mr^{2}
The various forces acting on the cylinder are shown in the following figure:
a = mg Sinθ / [m + (I/r^{2}) ]
= mg Sinθ / [m + {(1/2)mr^{2}/ r^{2}} ]
= (2/3) g Sin 30°
= (2/3) × 9.8 × 0.5 = 3.27 ms^{2}
^{}
(a) Using Newton’s second law of motion, we can write net force as:
f_{net} = ma
mg Sin 30°  f = ma
f = mg Sin 30°  ma
= 10 × 9.8 × 0.5  10 × 3.27
49  32.7 = 16.3 N
(b) During rolling, the instantaneous point of contact with the plane comes to rest. Hence, the work done against frictional force is zero.
(c) For rolling without skid, we have the relation:
μ = (1/3) tan θ
tan θ = 3μ = 3 × 0.25
∴ θ = tan^{1} (0.75) = 36.87°.
Question 7.32. Read each statement below carefully, and state, with reasons, if it is true or false;
(a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.
(b) The instantaneous speed of the point of contact during rolling is zero.
(c) The instantaneous acceleration of the point of contact during rolling is zero.
(d) For perfect rolling motion, work done against friction is zero.
(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.
Answer
Frictional force acts opposite to the direction of motion of the centre of mass of a body. In the case of rolling, the direction of motion of the centre of mass is backward. Hence, frictional force acts in the forward direction.
(b) True
Rolling can be considered as the rotation of a body about an axis passing through the point of contact of the body with the ground. Hence, its instantaneous speed is zero.
(c) False
This is becausehen a body is rolling, its instantaneous acceleration is not equal to zero. It has some value.
(d) True
This is because once the perfect rolling begins, force of friction becomes zero. Hence work done against friction is zero.
(e) True
This is because rolling occurs only on account of friction which is a tangential force capable of providing torque. When the inclined plane is perfectly smooth, the wheel will simply slip under the effect of its own weight.
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