A hollow sphere of radius 0.15 m, with rotational inertia I = 0.052 kg m2 about

Question

A hollow sphere of radius 0.15 m, with rotational inertia I = 0.052 kg m2 about a line through its center of mass, rolls without slipping up a surface inclined at 34° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 80 J.

(a) How much of this initial kinetic energy is rotational?

J

(b) What is the speed of the center of mass of the sphere at the initial position?

m/s

Now, the sphere moves 1.0 m up the incline from its initial position.

(c) What is its total kinetic energy now?

J

(d) What is the speed of its center of mass now?

m/s

Explanation / Answer

The rotational kinetic energy is E_rot = 1/2 I w^2, but because of no-slip w = v/R, so
E_rot = 1/2 I/R^2 v^2
The translational kinetic energy is E_tran = 1/2 m v^2, and the total kinetic energy is
E = 1/2 (I/R^2 + m) v^2

a) the fraction of rotational to total the kinetic energy is I/R^2 / ( I/R^2 + m ) = 1/(1+ mR^2 /I)
From the question, I have to assume the inner radius is (almost) equal to the outer radius, so I = mR^2, and the fraction = 1/(1+1) = 1/2

b)because from a) we know E_tran = E_rot, we have Etran= 40.0J. This equals Erot = 1/2 I/R^2 v^2, so v^2 = 2 Erot R^2/I = 2 * 40.0 J * 0.150^2 m^2 / (0.052 kg m^2)
and v = sqrt(2*40.0*0.150^2/0.052^2) = 25.80 m/s

c) the answer in c allows you to calculate the mass: 1/2 m v^2 = 40.0 J, so m = 80.0J/(25.8 m/s)^2 = 0.120 kg.
Then you can calculate the increase in potential energy going uphill:
U = m g h = 0.12 kg * 9.81 m/s^2 *1 m * sin(34) = 0.623 J.
This leaves 80.0 J -0.623 J = 79.37 J for total kinetic energy. Then use the derived formula for ttal kinetic energy in terms of v to calculate v.

V = sqrt(2*79.37/0.12) = 36.37 m/s

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