Physics Speed Question

Question

1.10 kg solid, ball of radius 0.140 m is released from rest at point A, its center of gravity a distance of 1.70 m above the ground. The ball rolls w/o slipping to the bottom of an incline and back up to point B where it is launched vertically into the air. The ball then rises to its maximum height h_max at point C.

a)At point B, find the ball's translational speed v_B ( m/s).

b)At point B, find the ball's rotational speed 𝜔_B (rad/s).

c)At point C, find the ball's rotational speed 𝜔_C (rad/s).

d)At point C, find the maximum height h_max of the ball's center of gravity (m).

height of A= 1.70m

height of B= 0.300m

Answer 1

Let

m = 1.10 kg be the ball's mass,

r = 0.140 m be the ball's radius,

I = 2mr²/5 be the moment of inertia of the ball,

h_A = 1.70 m be the height of point A,

h_B = 0.300 m be the height of point B,

h_max be the height of the ball at the maximal point C,

g = 9.81 m/s² be gravitational acceleration.

Given that no information is given about friction or resistance,

I am going to ignore them.

In that case all of the potential energy at point A is converted

at each of point B and C to the ball's potential energy plus its kinetic energy.

And its kinetic energy is the sum of its translational and its rotational energy.

Its potential energy at point B is mgh_B

Its translational kinetic energy is mv_B²/2

Its rotational kinetic energy is Iω_B²/2

As long as the ball is rotating without slipping

ω_B = v_B/r (1)

Using (1) we can rewrite the rotational kinetic energy as

I(v_B/r)²/2 = (2mr²/5)(v_B/r)²/2 = mv_B²/5

So the total energy at point B is

mgh_B + mv_B²/2 + mv_B²/5 = mgh_B + 7mv_B²/10

By conservation energy between points A and B

mgh_B + 7mv_B²/10 = mgh_A

Solving for v_B:

v_B² = 10g(h_A-h_B)/7 (2)

You can use equation (2) to answer question a),

and then equation (1) to answer question b).

Once the ball leaves the point B equation (1) no longer applies,

and by conservation of angular momentum the ball will retain its rotational speed;

that is ω_C = ω_B.

Therefore at the maximum height (where translational speed is 0), the ball's energy is

mgh_max + Iω_C²/2 = mgh_max + (2mr²/5)(v_B/r)²/2 = mgh_max + mv_B²/5 = mgh_max + 2mg(h_A-h_B)/7

This energy must equal the original potential energy:

mgh_max + 2mg(h_A-h_B)/7 = mgh_A (3)

Solving for h_max:

h_max = h_A - 2(h_A-h_B)/7

Please note that the answers to all the questions are independent of m and r;

that is, they are the same for any ball.

And answer to question d) is independent of g;

that is, the maximum height would be the same on any planet.

But it would not work in the absence of gravity, because we divided equation (3) by g,

which we would not be allowed to do if g were 0.