AP Physics B and C– Answer to Free-response Question on Circular Motion

In the post dated 22^nd January 2008, the following free-response question for practice was given to you:

A simple pendulum of length ℓ is suspended from the point O (fig.). The bob of the pendulum is a sphere of mass m and is initially at rest at A. An identical sphere S (not shown in the figure), which is a projectile with its trajectory in the plane of the figure, has its highest point at A. The projectile therefore collides (elastically) with the bob of the pendulum at A and makes it move with horizontal speed V₁ as shown. The acceleration due to gravity at the place is g.

(a) If the speed V₁ of the bob of the pendulum is just sufficient to make it travel along the circular path of radius ℓ, derive an expression for the speed V₂ of the bob at the highest point B of the circular path.

(b) Derive an expression for the kinetic energy of the bob in terms of m, ℓ and g just when it starts moving from A.

(c) Obtain an expression for the speed of the bob at C, when the string of the pendulum is horizontal.

(d) What was the kinetic energy of the projectile (sphere S) just before it collided with the bob? Give reason for your answer without writing theoretical steps.

(e) Briefly explain the nature of the motion of the projectile after hitting the bob.

As promised, I give below the answer:

(a) Since the bob just moves along the circular path, the centripetal force required for the circular motion at the highest point B is supplied by the weight mg alone. The tension in the string will be zero in this case so that we have, at B,

mV₂²/ℓ = mg

Therefore, V₂ = √(ℓg).

(b) The bob has gained gravitational potential energy mgh = mg×2ℓ = 2mℓg, on moving up from A to B, by losing an equal amount of kinetic energy. The kinetic energy of the bob at B is (½) mV₂² = (½) mℓg, since V₂ = √(ℓg).

Therefore, the kinetic energy (K_A ) of the bob at the lowest point A must be given by

K_A = 2mℓg +(½) mℓg = (5/2) mℓg

(c) The bob has gained gravitational potential energy mgℓ on moving up from A to C (through a height ℓ), by losing an equal amount of kinetic energy. The kinetic energy (K_C) of the bob at C is therefore gien by

K_C = (5/2) mℓg – mℓg = (3/2) mℓg

The speed V₃ of the bob at C is therefore given by

(½) mV₃² = (3/2) mℓg, from which V₃ = √(3ℓg)

(d) When a moving sphere suffers a head-on elastic collision with an identical stationary sphere, the entire kinetic energy of the moving sphere is transferred to the stationary sphere. (The velocities get interchanged).

The kinetic energy of the projectile (sphere S) just before hitting the bob was therefore equal to the kinetic energy of the bob just after the hit: (5/2) mℓg

(e) The projectile will momentarily come to rest on hitting the bob and will fall freely under gravity (vertically) thereafter.

Here are some additional points you must note in the present context:

Suppose in the above question, you were asked to find the tension T₃ in the string when the bob is in position C. You will then answer like this:

Since the string is horizontal when the bob is at C, the centripetal force required for the circular motion of the bob is supplied solely by the tension T₃ in the string and so we have

mV₃²/ℓ = T₃

Since V₃ = √(3ℓg) we obtain T₃ = 3mg.

If you are asked to find the tension T₁ in the string when the bob is in its lowest position A, you will answer like this:

In position A of the bob, the centripetal force is supplied by the resultant of two forces: (i) Tension T₁ which tries to pull the bob vertically upwards

(ii) Weight mg of the bob which acts vertically downwards

Therefore, we have mV₁²/ℓ = T₁ – mg so that

T₁ = mV₁²/ℓ + mg

Since the kinetic energy (½) mV₁² of the bob at A is (5/2) mℓg, we have

V₁ = √(5ℓg)

Substituting this value of V₁ in the above expression for T₁, we obtain

T₁ = 6mg

If the bob is just able to move along the vertical circle, the tension T₂ in the string when the bob is at the top (position B) of the circle is zero and hence the difference between the tensions is

T₁ – T₂ = 6mg

You can easily show that this difference is 6mg in the case of any vertical circle [even if V₁ is greater than √(5ℓg)].

You should remember that the minimum velocity required for the bob at the lowest point A so as to make the bob move along the circular path is √(5ℓg) = √(5Rg), using symbol R for the radius of the circle.

You will often find questions making use of this fact.