Question

A hollow sphere of radius 0.130 m, with rotational inertia I = 0.0931 kg·m2 about a line through its center of mass, rolls without slipping up a surface inclined at 22.7° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 110 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 1.20 m up the incline from its initial position, what are (c) its total kinetic energy and (d) the speed of its center of mass?

Answer 1

Answer:

a) $K_{rot} = 44.005\,J$, b) $v \approx 3.997\,\frac{m}{s}$, c) $K_{2} = 72.472\,J$, d) $v \approx 3.244\,\frac{m}{s}$

Explanation:

b) The kinetic energy for a rigid body is the sum of translational and rotational kinetic energy:

$K = K_{rot} + K_{tr}$

$K = \frac{1}{2}\cdot \left(I\cdot \omega^{2} + m\cdot v^{2}\right)$

$K = \frac{1}{2}\cdot \left[I\cdot \left(\frac{v}{R} \right)^{2} +m\cdot v^{2}\right]$

$K = \frac{1}{2}\cdot \left[\frac{I}{R^{2}}+\frac{3\cdot I}{2\cdot R^{2}} \right] \cdot v^{2}$

$K = \frac{5\cdot I}{4\cdot R^{2}}\cdot v^{2}$

The speed of the center of mass of the sphere at the initial position is:

$v = \sqrt{\frac{4\cdot K\cdot R^{2}}{5\cdot I} }$

$v = \sqrt{\frac{4\cdot (110\,J)\cdot (0.130\,m)^{2}}{5\cdot \left(0.0931\,kg\cdot m^{2} \right)} }$

$v \approx 3.997\,\frac{m}{s}$

a) The kinetic energy associated with the rotation is:

$K_{rot} = \frac{1}{2}\cdot \left(0.0931\,\frac{kg}{m^{2}}\right)\cdot \left(\frac{3.997\,\frac{m}{s} }{0.130\,m} \right)^{2}$

$K_{rot} = 44.005\,J$

It is 40 % of the total initial kinetic energy.

c) The total final kinetic energy is:

$K_{2} = 110\,J - \left[\frac{3\cdot \left(0.0931\,kg\cdot m^{2}\right)}{2\cdot (0.130\,m)^{2}} \right]\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.20\,m)\cdot \sin 22.7^{\textdegree}$

$K_{2} = 72.472\,J$

d) The speed of its center of mass is:

$v = \sqrt{\frac{4\cdot K\cdot R^{2}}{5\cdot I} }$

$v = \sqrt{\frac{4\cdot (72.472\,J)\cdot (0.130\,m)^{2}}{5\cdot \left(0.0931\,kg\cdot m^{2} \right)} }$

$v \approx 3.244\,\frac{m}{s}$