Many of life's failures are people who did not realize how close they were to success when they gave up.

– Thomas A.Edison

The following two questions will be beneficial for AP Physics B as well as C:

(1) A rain drop of mass *m* at a height *h* starts from rest and gets accelerated under gravity initially. Because of the viscosity of air, its motion is resisted and it finally falls down with a constant terminal velocity *v*_{T} and hits the ground. You may neglect the force of buoyancy. Assume that the acceleration due to gravity, *g* is constant and the rain drop moves vertically down over the entire path. The magnitude of the *average* viscous force on the rain drop during the *entire* downward trip is

(a) *mg*

(b) *mg* – (*mv*_{T}^{2}/*h*)

(c) *mg* + (*mv*_{T}^{2}/*h*)

(d) *mg* + (*mv*_{T}^{2}/2*h*)

(e)* mg* – (*mv*_{T}^{2}/2*h*) * *

The initial gravitational potential energy of the drop at height *h* is *mgh*. On reaching the ground, it has lost its gravitational potential energy but has gained kinetic energy equal to ½ *mv*_{T}^{2}.

The loss of energy because of air resistance is *mgh* – ½ *mv*_{T}^{2}.

If the magnitude of the *average* viscous force on the rain drop is *F*, we have

*Fh =** mgh* – ½ *mv*_{T}^{2} * *

Therefore, *F =** mg* – (*mv*_{T}^{2}/2*h*) * *

(2) In the above question what is the viscous force when the rain drop moves down with its terminal velocity?

(a) *mg* upwards

(b) zero

(c) *mg *downwards

(d) *mv*_{T}^{2}/*h*) upwards

(e)* mv*_{T}^{2}/*h*) downwards* *

When the rain drop moves down with its terminal velocity, the net force acting on it is zero. This means that the weight *mg* of the rain drop and the viscous force offered by air are equal and opposite. Therefore, the viscous force has magnitude *mg* and is directed *upwards*.

The following question involves resistive *torques* in rotational motion.

(3) One end of a light inextensible string is tied to a solid cylinder of mass *M* and radius *R*. The string is wound round the cylinder many times and a mass *m *is suspended from the free end of the string so that on releasing the mass, the cylinder rotates about its central axis. There exists an appreciable amount of friction between the cylinder and the axle about which the cylinder rotates. When the mass *m* which is initially at a height *h* from the ground (Fig.) is released from rest, it strikes the ground with speed *v*. How much energy is lost from the system for doing work against friction during the fall of the mass *m *through the height *h*?

(a) *mgh *– (½ *mv*^{2}*+* ¼ *Mv*^{2})

(b) *mgh *– ½ *mv*^{2}

(c) *mgh *– ½ (*m + M*)*v*^{2}

(d) *mgh *– (½ *mv*^{2}*+* ½ *MR*^{2}*v*^{2})

(e)* mgh *– (½ *mv*^{2}*+* ¼ *MR*^{2}*v*^{2})

The system acquires kinetic energy at the cost of the gravitational potential energy of the suspended mass *m*.

The initial energy *E*_{1}* *of the system is given by

*E*_{1 }= *mgh*.

The final energy *E*_{2 }of the system at the moment the mass *m* hits the ground is the sum of the translational kinetic energy of the mass *m *and the rotational kinetic energy of the cylinder:

*E*_{2} = ½ *mv*^{2} + ½ (*MR*^{2}/2)×(*v*^{2}*/R*^{2})

[The second term on the right hand side is ½ *Iω*^{2} where *I* is the moment of inertia of the cylinder (*MR*^{2}/2) and *ω* is its angular velocity (*v/R*) when the mass *m *strikes the ground].

Therefore, *E*_{2} = ½ *mv*^{2}* +* ¼ * Mv*^{2}

The work (*W*) done by the system against frictional forces is given by

*W = E*_{1 }– *E*_{2} = *mgh *– (½ *mv*^{2}*+* ¼ *Mv*^{2})

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