Question

Question: A hollow sphere of radius 0.15 m, with rotational inertia $\mathit{I}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{040}\mathbf{}{\mathbf{\text{kgm}}}^{\mathbf{\text{2}}}$ about a line through its centre of mass, rolls without slipping up a surface inclined at 30⁰to the horizontal. At a certain initial position, the sphere’s total kinetic energy is 20J. (a) How much of this initial kinetic energy is rotational?

(b) What is the speed of the centre of mass of the sphere at the initial position? When the sphere has moved 1.0 mup the incline from its initial position,

(c) what is its total kinetic energy (d) What are the speed of its centre of mass?

Short answer

Answer

1. The fraction of the initial kinetic energy which is rotational is 0.4.
2. The speed of the center of mass of the sphere at the initial position is 3.0 m/s
3. The total kinetic energy of the sphere when it has moved 1.0 up the incline from its initial position is 6.9 J .
4. The speed of the center of mass of the sphere when it has moved up the incline from its initial position is 1.8 m/s

Step-by-step solution

Given

1. The radius of the hollow sphere is R =0.15 m
2. The rotational inertia of the hollow sphere is$\text{I}=0.04\text{kg}.{\text{m}}^2.$
3. The angle made by the inclined surface with the horizontal is$\theta ={30}^0$
4. The sphere’s total KE at certain initial position is 20J

To understand the concept

From the formula for rotational inertia of hollow sphere, we can find its mass. Using this value in formula for total K.E, we can find the velocity of the sphere. Then putting it into the formula for rotational and translational K.E and taking the ratio of rotational K.E to the total K.E, we can find how much fraction of the total K.E is the rotational K.E.

Then using conservation of the energy, we can find the total K.E of the sphere when it has moved up the incline from its initial position. Using the relation between total K.E and v for hollow sphere, we can find the final velocity of the center of mass of the sphere.

The law of conservation of energy is,

$\mathbf{U}\mathbf{+}{\mathbf{K}}_{\mathbf{tran}}\mathbf{+}{\mathbf{K}}_{\mathbf{rot}}\mathbf{=}\mathbf{constant}$

Here U is the potential energy, K_tanis the translational kinetic energy and K_rot is the rotational kinetic energy of the system.

(a) Calculate how much of the initial kinetic energy is rotational  

The moment of inertia of the hollow sphere is,

$I=\frac{2}{3}M{R}^2$

So, the mass of the hollow sphere is

$$\begin{gathered} M=\frac{3I}{2R^2} \\ =\frac{3\left(0.04{\text{kg.m}}^{\text{2}}\right)}{2{\left(0.15\text{m}\right)}^2} \\ =2.67\text{kg} \end{gathered}$$

(b) Calculate the speed of the center of mass of the sphere at the initial position

Translational kinetic energy of the object can be written as,

$K_{tran}=\frac{1}{2}m{v}^2$

v is the velocity with which center of mass of the sphere is moving.

Rotational kinetic energy can be written as,

$K_{rot}=\frac{1}{2}I{\omega }^2$

So, the fraction of the initial K.E which is rotational is,

$\frac{K_{rot}}{\left(K_{tran}\right)+\left(K_{rot}\right)}=\frac{\frac{1}{2}I{\omega }^2}{\left(\frac{1}{2}M{v}^2\right)+\left(\frac{1}{2}I{\omega }^2\right)}$

But,

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

Therefore,

$\frac{K_{rot}}{\left(K_{tran}\right)+\left(K_{rot}\right)}=\frac{\frac{1}{2}I\left(\frac{v^2}{R^2}\right)}{\left(\frac{1}{2}M{v}^2\right)+\left(\frac{1}{2}I\left(\frac{v^2}{R^2}\right)\right)}$

We have given that

$$\begin{gathered} K=\left(K_{tran}\right)+\left(K_{rot}\right) \\ =20\text{J} \\ \left(\frac{1}{2}M{v}^2\right)+\left(\frac{1}{2}\frac{2}{3}M{R}^2\left(\frac{v^2}{R^2}\right)\right)=20\text{J} \\ \left(\frac{1}{2}M+\frac{1}{3}M\right)v^2=20\text{J} \\ \frac{5}{6}M{v}^2=20\text{J} \\ {v}^2=\frac{20\text{J}}{\frac{5}{6}\left(2.7\text{kg}\right)} \\ {\text{v}}^2=8.89\frac{{\text{m}}^2}{{\text{s}}^2} \end{gathered}$$

Therefore,

$$\begin{gathered} \frac{K_{rot}}{\left(K_{tran}\right)+\left(K_{rot}\right)}=\frac{\frac{1}{2}I\left(\frac{v^2}{R^2}\right)}{20\text{J}} \\ =\frac{\frac{1}{2}\left(0.04{\text{kg.m}}^{\text{2}}\right)\left(\frac{\left(8.88m^2/s^2\right)}{{\left(0.15\text{m}\right)}^2}\right)}{20\text{J}} \\ =0.394 \\ =0.4 \end{gathered}$$

Therefore, 0.4 of the initial K.E is rotational.

(c) Calculate the total kinetic energy  

In part (a), we have got

$$\begin{gathered} v^2=8.88{\left(\text{m/s}\right)}^2 \\ v=2.98\text{m/s} \\ =3.0\text{m/s} \end{gathered}$$

Therefore,the speed of the center of mass of the sphere at initial position is 3.0 m/s.

If the sphere is moved up the incline the change in height will be-

$$\begin{gathered} h=1\times \text{Sin}30° \\ =0.5\text{m} \end{gathered}$$

According to the law of conservation of the total mechanical energy,

$K_i+U_i=K_f+U_f$

Let the initial value of the potential energy be zero, then the final potential energy will be-

$U_f=Mgh$

For the given values of the kinetic and potential energy-

$$\begin{gathered} 20\text{J}+0=K_f+\left(2.67\text{kg}\right)\left(9.8\right)\left(0.5\text{h}\right) \\ {K}_f=6.9\text{J} \end{gathered}$$

Therefore, the total K.E of the sphere when it has moved a meter up the incline from its initial position is 6.9 J.

(d) Calculate the speed of its center of mass

From part (a) and part (c), we have

$$\begin{gathered} K_f=\frac{5}{6}M{v}^2 \\ {v}^2=\frac{6.9\text{J}}{\frac{5}{6}\left(2.67\text{kg}\right)} \\ {v}^2=3.1m/s^2 \\ v=1.8\text{m/s} \end{gathered}$$

Therefore, the speed of the center of mass of the sphere when it has moved 1.0 m up the incline from its initial position is 1.8 m/s.