Question 70: Centre of Mass and Rotational Motion

Question

An object of uniform density rolls up the curved path with the initial velocity $v_{\mathrm{o}}$ as shown in the figure. If the maximum height attained by an object is $\frac{7 v_0^2}{10 \mathrm{~g}}$ ( $\mathrm{g}=$ acceleration due to gravity), the object is a $\_\_\_\_$ .

Accepted Answer

At the bottom where the object is projected,
Total kinetic energy (K. E.total ) is the sum of translational K.E. (K. E.t ) and rotational K.E. (K. E. ${ }_{\mathrm{r}}$ ):
$$ \mathrm{K} . \mathrm{E}_{. 	ext {total }}=\frac{1}{2} \mathrm{mv}_0^2+\frac{1}{2} \mathrm{I} \omega^2 $$
Where I is the moment of inertia and m is the mass of object.
For pure rolling, the relationship between linear velocity ( $\mathrm{v}_0$ ) and angular velocity ( $\omega$ ) is $\mathrm{v}_0=\mathrm{R} \omega \Rightarrow \omega=\frac{\mathrm{v}_0}{\mathrm{R}}$.
The moment of inertia can be written as $\mathrm{I}=\mathrm{mk}^2$, where k is the radius of gyration.
$$ \mathrm{K} . \mathrm{E}_{\cdot 	ext { total }}=\frac{1}{2} \mathrm{mv}_0^2+\frac{1}{2}\left(\mathrm{mk}^2ight)\left(\frac{\mathrm{v}_0}{\mathrm{R}}ight)^2 $$
$$\Rightarrow $$ $$ \mathrm{K} . \mathrm{E}_{	ext {total }}=\frac{1}{2} \mathrm{mv}_0^2\left(1+\frac{\mathrm{k}^2}{\mathrm{R}^2}ight) $$
At the maximum height (h), the object momentarily comes to rest. All its kinetic energy is converted into gravitational potential energy (U):
$$ \mathrm{U}=\mathrm{mgh} $$
The maximum height is given as $\mathrm{h}=\frac{7 \mathrm{v}_0^2}{10 \mathrm{~g}}$
$$ 	ext { So, } U=m g\left(\frac{7 v_0^2}{10 g}ight)=\frac{7}{10} \mathrm{mv}_0^2 $$
Using conservation of energy
$$ 	ext { K. } \mathrm{E}_{	ext {total }}=\mathrm{U} $$
$$\Rightarrow $$ $\frac{1}{2} m v_0^2\left(1+\frac{k^2}{R^2}ight)=\frac{7}{10} m v_0^2$
$$\Rightarrow $$ $\frac{1}{2}\left(1+\frac{\mathrm{k}^2}{\mathrm{R}^2}ight)=\frac{7}{10}$
$$\Rightarrow $$ $1+\frac{\mathrm{k}^2}{\mathrm{R}^2}=\frac{14}{10}=1.4$
$$\Rightarrow $$ $$ \frac{\mathrm{k}^2}{\mathrm{R}^2}=1.4-1=0.4=\frac{2}{5} \Rightarrow \mathrm{k}^2=\frac{2}{5} \mathrm{R}^2 $$
So, the moment of inertia of object is, $\mathrm{I}=\mathrm{mk}^2=\frac{2}{5} \mathrm{mR}^2$
For a solid sphere, $\mathrm{I}=\frac{2}{5} \mathrm{mR}^2 \Rightarrow \frac{\mathrm{k}^2}{\mathrm{R}^2}=\frac{2}{5}$.
For a solid cylinder/disc, $\mathrm{I}=\frac{1}{2} \mathrm{mR}^2 \Rightarrow \frac{\mathrm{k}^2}{\mathrm{R}^2}=\frac{1}{2}$.
For a ring/hollow cylinder, $\mathrm{I}=\mathrm{mR}^2 \Rightarrow \frac{\mathrm{k}^2}{\mathrm{R}^2}=1$.
Therefore, the object is a solid sphere.
Hence, the correct option is (D).

Answer

solid cylinder

Answer

ring

Answer

disc

Answer

solid sphere

Centre of Mass and Rotational Motion

An object of uniform density rolls up the curved path with the initial velocity v_{\mathrm{o}} as shown in the figure. If the maximum height attained by an object is \frac{7 v_0^2}{10 \mathrm{~g}} ( \mathrm{g}= acceleration due to gravity), the object is a \_\_\_\_ .

Choose the correct answer: