Geometric centre; No
The centre of mass (C.M.) is a point where the mass of a body is supposed to be concentrated. For the given geometric shapes having a uniform mass density, the C.M. lies at their respective geometric centres.
The centre of mass of a body need not necessarily lie within it. For example, the C.M. of bodies such as a ring, a hollow sphere, etc., lies outside the body.
Let m and r be the respective masses of the hollow cylinder and the solid sphere.
The moment of inertia of the hollow cylinder about its standard axis,II = mr2
The moment of inertia of the solid sphere about an axis passing through its centre, III = 2/5 mr2
We have the relation:
τ = I α
Where, α = Angular acceleration
τ = Torque
I = Moment of inertia
For the hollow cylinder, τI = II αI
For the solid sphere, τII = III αII
As an equal torque is applied to both the bodies, τI = τ2
∴ αII / αI = II / III = mr2 / 2/5 mr2 = 2/5
αII > αI .... (i)
Now, using the relation:
ω = ω0 + αt
Where, ω0 = Initial angular velocity
t = Time of rotation
ω = Final angular velocity
For equal ω0 and t, we have:
ω ∝ α … (ii)
From equations (i) and (ii), we can write:
ωII > ωI
Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.
Mass of the cylinder, m= 20 kg
Angular speed, ω = 100 rad s-1
Radius of the cylinder, r= 0.25 m
The moment of inertia of the solid cylinder:
I = mr2 / 2
= 1/2 x 20 x (0.25)2
= 0.625 kgm2
∴Kinetic energy = 1/2 I ω2
= 1/2 x 6.25 x 1002 = 3125 J
∴Angular momentum, L= Iω = 6.25 × 100 = 62.5 Js
(a) 100 rev/min
Initial angular velocity, ω1= 40 rev/min
Final angular velocity = ω2
The moment of inertia of the boy with stretched hands = I1
The moment of inertia of the boy with folded hands = I2
The two moments of inertia are related as:
I2 = 2/5 I1
Since no external force acts on the boy, the angular momentum L is a constant.
Hence, for the two situations, we can write:
I2 ω2 = I1 ω1
ω2 = I1 ω1 / I2
= I1 / 2/5 I2 x 40 = 5/2 x 40
= 100 rev/min
(b) Final K.E. = 2.5 Initial K.E.
Final kinetic rotation,EF = 1/2 I2 ω22
Initial kinetic rotation, EI = 1/2 I1 ω12
EF / EI = 1/2 I2 ω22 / 1/2 I1 ω12
= 2/5 I1 / I1 x 1002 / 402
= 5/2 = 2.5
∴ EF = 2.5 EI
The increase in the rotational kinetic energy is attributed to the internal energy of the boy.
Mass of the hollow cylinder, m = 3 kg
Radius of the hollow cylinder, r = 40 cm = 0.4 m
Applied force, F = 30 N
The moment of inertia of the hollow cylinder about its geometric axis:
I = mr2
= 3 × (0.4)2 = 0.48 kg m2
Torque, τ = F x r
= 30 × 0.4 = 12 Nm
For angular acceleration α, torque is also given by the relation:
τ = Iα
α = τ / I = 12 / 0.48
= 25 rad s-2
Linear acceleration = rα = 0.4 × 25 = 10 m s–2
Angular speed of the rotor, ω = 200 rad/s
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 × 103 = 36 kW
Hence, the power required by the engine is 36 kW.
Radius of the hoop, r = 2 m
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 KE + Rotational KE
Er = 1/2 mv2 + 1/2 I ω2
Moment of inertia of the hoop about its centre, I = mr2
Er = 1/2 mv2 + 1/2 (mr2) ω2
But we have the relation, v = rω
∴ Er = 1/2 mv2 + 1/2 mr2 ω2
∴ Er = 1/2 mv2 + 1/2 mv2
∴ Er = mv2
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 = mv2 = 100 × (0.2)2 = 4 J
A torque is required to roll the given disc. As per the definition of torque, the rotating force should be tangential to the disc. Since the frictional force at point B is along the tangential force at point A, a frictional force is required for making the disc roll.
(a) Force of friction acts opposite to the direction of velocity at point B. The direction of linear velocity at point B is tangentially leftward. Hence, frictional force will act tangentially rightward. The sense of frictional torque before the start of perfect rolling is perpendicular to the plane of the disc in the outward direction.
(b) Since frictional force acts opposite to the direction of velocity at point B, perfect rolling will begin when the velocity at that point becomes equal to zero. This will make the frictional force acting on the disc zero.
No change
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.
(a) False 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 When a body is rolling, its instantaneous acceleration is not equal to zero. It has some value.
(d) True When perfect rolling begins, the frictional force acting at the lowermost point becomes zero. Hence, the work done against friction is also zero.
(e) True The rolling of a body occurs when a frictional force acts between the body and the surface. This frictional force provides the torque necessary for rolling. In the absence of a frictional force, the body slips from the inclined plane under the effect of its own weight.