Question

In the given figure, two wheels P and Q are connected by a belt B. The radius of P is three times as that of Q. In case of same rotational kinetic energy, the ratio of rotational inertias $(\frac{I_1}{I_2})$ will be x : 1. The value of x will be ________ . 

 

Hint:

For pulleys or wheels connected by a non-slipping belt, their linear (tangential) speeds at the point of contact are equal. This allows you to relate their angular speeds based on their radii ($\omega_1 r_1 = \omega_2 r_2$).

Answer 1

Correct Answer: 9

Let the radius of wheel Q be $R_Q = R$.
The radius of wheel P is three times that of Q, so $R_P = 3R$.
Since the two wheels are connected by a belt that does not slip, the tangential speed at the rim of both wheels must be the same.
$v_P = v_Q$.
The relationship between tangential speed (v), angular speed ($\omega$), and radius (r) is $v = \omega r$.
$\omega_P R_P = \omega_Q R_Q$.
Substituting the radii:
$\omega_P (3R) = \omega_Q (R)$.
$3\omega_P = \omega_Q$.
The problem states that the rotational kinetic energies of the two wheels are the same.
$K_P = K_Q$.
The formula for rotational kinetic energy is $K = \frac{1}{2}I\omega^2$.
$\frac{1}{2}I_P \omega_P^2 = \frac{1}{2}I_Q \omega_Q^2$.
Substitute $\omega_Q = 3\omega_P$ into the equation:
$I_P \omega_P^2 = I_Q (3\omega_P)^2$.
$I_P \omega_P^2 = I_Q (9\omega_P^2)$.
Cancel $\omega_P^2$ from both sides (assuming the wheels are rotating).
$I_P = 9 I_Q$.
The question asks for the ratio of rotational inertias, let's assume $I_1 = I_P$ and $I_2 = I_Q$.
$\frac{I_1}{I_2} = \frac{I_P}{I_Q} = \frac{9I_Q}{I_Q} = 9$.
The ratio is 9 : 1. Therefore, the value of x is 9.