System of Particles and Rotational Motion - Result Question 58
62. A small mass attached to a string rotates on frictionless table top as shown. If the tension in the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2 , the kinetic energy of the mass will
[2011M]
(a) remain constant
(b) increase by a factor of 2
(c) increase by a factor of 4
(d) decrease by a factor of 2
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Answer:
Correct Answer: 62. (c)
Solution:
(c) K.E. $=\frac{L^{2}}{2 I}$
The angular momentum $L$ remains conserved about the centre.
That is, $L=$ constant.
$I=mr^{2}$
$\therefore$ K.E. $=\frac{L^{2}}{2 mr^{2}}$
In 2 nd case, K.E. $=\frac{L^{2}}{2(mr^{\prime 2})}$
But $r^{\prime}=\frac{r}{2}$
$\therefore K . E^{\prime}=\frac{L^{2}}{2 m \cdot \frac{r^{2}}{4}}=\frac{4 L^{2}}{2 mr^{2}} \Rightarrow$ K.E.’ $=4$ K.E.
$\therefore$ K.E. is increased by a factor of 4 .
When no torque is applied angular momentum remains constant.
$I_1 \omega_1=I_2 \omega_2 \Rightarrow \frac{\omega_2}{\omega_1}=\frac{I_1}{I_2}$
$\therefore \frac{E_2}{E_1}=\frac{\frac{1}{2} I_2 \omega_2^{2}}{\frac{1}{2} I_1 \omega_1^{2}}=\frac{I_2}{I_1}(\frac{I_1}{I_2})^{2}=\frac{I_2}{I_2}$
