Question: Non-uniform ball. In Figure, a ball of mass M and radius R rolls smoothly from rest down a ramp and onto a circular loop of radius 0.48 m The initial height of the ball is h =0.36m. At the loop bottom, the magnitude of the normal force on the ball is 2.00 mg. The ball consists of an outer spherical shell (of a certain uniform density) that is glued to a central sphere (of a different uniform density). The rotational inertia of the ball can be expressed in the general form $\mathit{I}\mathbf{=}\mathit{\beta }\mathit{M}{\mathit{R}}^{\mathbf{2}}$, but$\mathit{\beta }$is not 0.4as it is for a ball of uniform density. Determine $\mathit{\beta }$.

1. The mass of the ball is M
2. The radius of the ball is R
3. The radius of the circular loop is r = 0.48 m
4. The initial height of the ball is h =0.36 m
5. The magnitude of the normal force acting on the ball at the bottom of the loop is N = 2.00 mg
6. The expression for rotational inertia$I=\beta M{R}^2$

According to the rule of conservation of energy, energy can only be transformed from one form of energy to another and cannot be created or destroyed.

Applying Newton’s second law to the ball at the bottom of the trajectory, get a force equation. Then by putting values in it, get an expression for ,from which angular speed of the ball can be found. Then putting it in the equation obtained from the law of conservation of energy, get an expression for M.I of the ball about the center of mass. After comparing with the given expression for M.I , get the value of$\beta$.

Formulae:

The law conservation of energy is,

Initial total energy of the system = Final total energy of the system

$\omega =\frac{v}{R}$

When the ball reaches the bottom of the circular trajectory, the necessary centripetal force is provided by the difference between the normal force and the weight of the ball. So we can write

$$\begin{gathered} N-Mg=\frac{M{v}^2}{r} \\ N=Mg+\frac{M{v}^2}{r} \end{gathered}$$

Substituting the value of N from the given,

$$\begin{gathered} 2Mg=Mg+\frac{M{v}^2}{r} \\ {v}^2=gr \\ {v}^2=9.8m/s^2 \left(0.48 \text{m}\right) \\ =4.7m^2/s^2 \end{gathered}$$

Relationship between linear and angular velocity is$v=R\omega$

The angular speed of the ball of radius is R

$$\begin{gathered} {\omega }^2=\frac{v^2}{R^2} \\ =\frac{gr}{R^2} \end{gathered}$$

The law of energy conservation gives

Initial total energy of the system = Final total energy of the system

$$\begin{gathered} Mgh=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2 \\ \Rightarrow Mgh=\frac{1}{2}M{v}^2+\frac{1}{2}I\left(\frac{v^2}{R^2}\right) \\ \Rightarrow Mgh-\frac{{\displaystyle 1}}{{\displaystyle 2}}M{v}^2=\frac{{\displaystyle 1}}{{\displaystyle 2}}I\left(\frac{{\displaystyle v^2}}{{\displaystyle R^2}}\right) \\ \Rightarrow I=\frac{{\displaystyle 2Mh{R}^2}}{{\displaystyle r}}-M{R}^2 \\ \Rightarrow I=\left(\frac{{\displaystyle 2h}}{{\displaystyle r}}-1\right)M{R}^2 \\ \Rightarrow I=\left(\frac{{\displaystyle 2\left(0.36 \text{m}\right)}}{{\displaystyle 0.48 \text{m}}}-1\right)M{R}^2 \\ \Rightarrow I=0.50M{R}^2 \end{gathered}$$

We have,

$I=\beta M{R}^2$

Comparing these two equations, we get

$\beta =0.50$

Hence, the value of $\beta is0.50$