A hollow sphere of radius 0.240 m, with rotational inertia I = 0.0969 kg·m2 abou

Question

A hollow sphere of radius 0.240 m, with rotational inertia I = 0.0969 kg·m2 about a line through its center of mass, rolls without slipping up a surface inclined at 24.2° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 29.0 J. (a) How much of this initial kinetic energy is rotational? (b) What is the speed of the center of mass of the sphere at the initial position? When the sphere has moved 0.950 m up the incline from its initial position, what are (c) its total kinetic energy and (d) the speed of its center of mass?

Explanation / Answer

Here, 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.

Further, 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= 29.0/2 = 14.5 J.

This equals Erot = 1/2 I/R^2 v^2, so v^2 = 2 Erot R^2/I = 2 * 14.5 J * 0.240^2 m^2 / (0.0969 kg m^2)
and v = sqrt(2*14.5*0.240^2/0.0969^2) = 13.34 m/s

(c) the velocity calculated above allows you to calculate the mass:

=> 1/2 m v^2 = 14.5 J, so m = 29.0J/(13.34 m/s)^2 = 0.163 kg.
Then you can calculate the increase in potential energy going uphill:
U = m g h = 0.163 kg * 9.81 m/s^2 * 0.950 m * sin(24.2) = 0.623 J.
This leaves 29.0 J - 0.623 J = 28.377 J for total kinetic energy.

(d) Now,  E = 1/2 (I/R^2 + m) v^2 = 28.377

=> (I/R^2 + m) v^2 = 56.754

=> (0.0969/0.240^2 + 0.163)v^2 = 55.754

=> (1.6823 + 0.163)v^2 = 55.754

=> v^2 = 30.214

=> v = 5.50 m/s

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