Question 410: Centre of Mass and Rotational Motion

Question

Two discs of moment of inertia $$I_1=4 \mathrm{~kg} \mathrm{~m}^2$$ and $$I_2=2 \mathrm{~kg} \mathrm{~m}^2$$, about their central axes & normal to their planes, rotating with angular speeds $$10 \mathrm{~rad} / \mathrm{s}$$ & $$4 \mathrm{~rad} / \mathrm{s}$$ respectively are brought into contact face to face with their axes of rotation coincident. The loss in kinetic energy of the system in the process is _________ J.

Accepted Answer

To find the loss in kinetic energy when two spinning discs are brought together, we use the principle of conservation of angular momentum and the formula for kinetic energy. Here's how:
First, because angular momentum before and after they touch must be the same, we have:
$$I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega_0$$
where:
$$I_1$$ and $$I_2$$ are the moments of inertia for the two discs.
$$\omega_1$$ and $$\omega_2$$ are their angular speeds before contact.
$$\omega_0$$ is their common angular speed after contact.
Plugging in the given values, we find that:
$$\omega_0 = 8 \mathrm{rad/s}$$
Next, to calculate the loss in kinetic energy, we first find the total kinetic energy before and after they touch:
Before: $$E_1 = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 = 216 \,\mathrm{J}$$
After: $$E_2 = \frac{1}{2}(I_1 + I_2) \omega_0^2 = 192 \,\mathrm{J}$$
The loss in kinetic energy ($$\Delta E$$) is the difference:
$$\Delta E = E_1 - E_2 = 24 \,\mathrm{J}$$
So, when the two discs are brought together, the system loses 24 J of kinetic energy.

Centre of Mass and Rotational Motion

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