Two identical small balls, each having a mass of 1/2 kg , are attached to the ends of a light rod that is 50 cm long to form a dumb-bell. This dumb-bell rotates at an angular speed of 10 rad/s about a fixed axis passing

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Correct answer: D. 12 rad/s

The moment of inertia of the dumb-bell system about the axis of rotation is given by:

I = 2mr^2

Substituting

m = 0.5 \text{ kg}

and

r = \dfrac{0.5}{2} = 0.25 \text{ m}

:

I = 2 \times 0.5 \times (0.25)^2 = 0.0625 \text{ kg m}^2

The initial angular momentum of the system is:

L_i = I\omega_0 = 0.0625 \times 10 = 0.625 \text{ kg m}^2/\text{s}

The torque produced by the impulsive force about the centre is:

\tau = F \times r = 5.0 \times 0.25 = 1.25 \text{ N m}

The angular impulse applied to the system is:

\Delta L = \tau \Delta t = 1.25 \times 0.10 = 0.125 \text{ kg m}^2/\text{s}

Since the force is applied in the direction of velocity, the angular momentum increases. The final angular momentum is:

L_f = L_i + \Delta L = 0.625 + 0.125 = 0.750 \text{ kg m}^2/\text{s}

The new angular velocity is:

\omega_f = \dfrac{L_f}{I} = \dfrac{0.750}{0.0625} = 12 \text{ rad/s}

Two identical small balls, each having a mass of 1/2 kg , are...

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