(1) A car moves forward on level road with uniform velocity of 4 ms^{–1}. If there is no slipping of the tyres, the velocities (with respect to the road) of the points at the bottom, centre and the top of a tyre are respectively

(a) 4 ms^{–1}, 4 ms^{–1 }and 4 ms^{–1}

(b) zero, 4 ms^{–1}and 8 ms^{–1}

(c) 4 ms^{–1}, zero^{ }and 8 ms^{–1}

(d) 8 ms^{–1}, 4 ms^{–1}and 8 ms^{–1}

(e) 8 ms^{–1}, 4 ms^{–1} and zero

*rotation*, points at the bottom of the wheel move

*backward*with speed 4 ms

^{–1}where as points at the top move

*forward*with speed 4 ms

^{–1}. The resultant speed of points at the bottom is therefore zero (4 ms

^{–1 }– 4 ms

^{–1})

^{ }where as that of points at the top is 8 ms

^{–1 }(4 ms

^{–1 }+ 4 ms

^{–1}). The linear velocity of the centre of the wheel because of its

*rotation*is zero and hence the velocity of 4 ms

^{–1}due to the translational motion is preserved at the central point of the wheel.

^{ }The correct option is (b).

(2) Three solid spheres of masses 10 kg, 8 kg and 2 kg are connected by two springs of negligible mass and force constants 10 Nm^{–1} and 12 Nm^{–1} as shown. The spheres are at rest on a horizontal frictionless surface so that their centres are along the *x*-axis. An impulsive force is applied on the 10 kg mass so that it starts moving with a velocity of 2 ms^{–1} along the positive *x*-direction. The velocity of the centre of mass of the system of spheres is

(a) 0.5 ms^{–1}

(b) 0.8 ms^{–1}

(c) 1 ms^{–1}

(d) 1.6 ms^{–1}

(e) 2 ms^{–1}

Some of you will be confused by this simple problem. You should remember that the forces developed in the springs are *internal* forces in the system and hence they will not affect the velocity of the centre of mass. The law of conservation of momentum is obeyed and we can equate the initial momentum to the final momentum:

10×2 + 8×0 + 2×0 = (10+8+2)*v* where *v* is the velocity of the centre of mass.

(3) A horizontal turn table of mass *M *in the form of a uniform circular disc of radius *R *is rotating about a central vertical axis with uniform angular velocity *ω*. The friction at the axis is negligible. When two equal masses *m *and *m* are gently placed symmetrically on either side of the centre at distance *R/*2 (Fig), the angular velocity of the turn table is reduced to* ω/*2. Then each mass *m *is equal to

(a) *M/*4

(b)* M/*2

(c) *M*

(d) 3*M/*2

(e) 2*M*

This question is meant for checking your understanding of the law of conservation of angular momentum and for testing whether you can apply it in situations where it is needed.

The angular momentum is conserved in the absence of external torque. The angular momentum of the system before and after placing the two equal masses is the same so that we have

(*MR*^{2}/2)* ω = *[(*MR*^{2}/2) + 2*m *(*R/*2)^{2}] *ω/*2

[Note that *MR*^{2}/2 is the moment of inertia of the disc about the central axis and 2*m *(*R/*2)^{2} is the total moment of inertia of the two masses *m *and *m* about the same axis which is distant *R/*2 from them].

From the above equation *MR*^{2}/4 = *mR*^{2}/4 so that *m = M*

(4) A solid sphere S_{1} spinning about its own central axis with angular velocity *ω* moves along a frictionless horizontal surface and undergoes a head-on elastic collision with an identical solid sphere S_{2} at rest on the surface. After the collision the spin angular velocities of S_{1} and S_{2} will be respectively

(a)* ω/*2 and* ω/*2

(b) *ω *and zero

(c) zero and *ω*

(d) zero and 2*ω*

(e) 2*ω *and zero

Since the line of action of the force exerted by S_{1} on S_{2} passes through the centre of S_{2}, there is no torque on S_{2}. Its spin angular momentum after the collision is therefore *zero* (same as that before the collision). Since the angular momentum of the system is to be conserved, the angular momentum of S_{1} also is unchanged by the collision. Hence its spin angular velocity after the collision is *ω*. The correct option is (b).

(5) A car is decelerating uniformly from velocity *v*_{1} to velocity* v*_{2} while getting displaced by *s*. If the diameter of the wheel is *d*, the angular acceleration of the wheel is

(a) (*v*_{1}^{2} – *v*_{2}^{2})/*ds*

(b) (*v*_{2} – *v*_{1})/*ds*

(c) (*v*_{1} – *v*_{2})/*ds*

(d) (*v*_{2}^{2} – *v*_{1}^{2})/*ds*

(e) 2(*v*_{2}^{2} – *v*_{1}^{2})/*ds*

The acceleration ‘*a*’* *of the car is given by

*v*_{2}^{2} = *v*_{1}^{2} + 2*as*

[We have applied the equation of linear motion, *v*^{2} = *v*_{0}^{2} + 2*as* (or, *v*^{2} = *u*^{2} + 2*as*)].

Therefore, *a =* (*v*_{2}^{2} – *v*_{1}^{2})/2*s*.

The angular acceleration ‘α’ of the wheel is given by

α = *a/r* where *r* is the radius of the wheel (which is equal to *d/*2).

Therefore, α = [(*v*_{2}^{2} – *v*_{1}^{2})/2*s*]/(*d*/2) = (*v*_{2}^{2} – *v*_{1}^{2})/*ds*

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