A uniform square plate is placed on a horizontal floor. When it is given an angular velocity o about a vertical axis through one of its corners as shown in figure-I, it takes time t1 to come to a complete stop. Now the same square plate is given the same angular velocity to rotate about another vertical axis through its centre as shown in figure-II. How long will it take to come to a complete stop now?

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The time it takes for the square plate to come to a complete stop depends on the moment of inertia, which is different for rotation about a corner versus rotation about the center.

The moment of inertia (I) for a square plate of mass (m) and side (a) rotating about an axis through one of its corners is given by I = (m*a^2)/6.

The moment of inertia for the same square plate rotating about an axis through its center is given by I = (m*a^2)/12.

The angular velocity (ω) is the same in both cases.

The rotational kinetic energy (K.E) is given by K.E = 0.5Iω^2.

Assuming no external torques, the rotational kinetic energy will be dissipated at the same rate in both cases due to friction. Therefore, the time (t) it takes for the plate to come to a stop is proportional to its initial rotational kinetic energy.

So, t1/t2 = K.E1/K.E2 = I1/I2 = [(ma^2)/6] / [(ma^2)/12] = 2.

Therefore, it will take twice as long for the square plate to come to a complete stop when it is rotating about an axis through its center compared to when it is rotating about an axis through one of its corners.

To answer this question, we need to understand the concept of rotational inertia or moment of inertia, which is the rotational equivalent of mass in linear motion. The moment of inertia depends on the mass of the object and its distribution of mass around the axis of rotation.

For a square plate of side 'a' and mass 'm', the moment of inertia when it is rotating about an axis through one of its corners (diagonal as axis) is given by I1 = (m*a^2)/6.

When the same plate is rotating about an axis through its center, the moment of inertia is given by I2 = (m*a^2)/12.

The angular momentum (L) of the plate is conserved in both cases, and is given by L = I*ω, where ω is the angular velocity.

Since the angular velocity is the same in both cases, we can write L1 = L2, or I1ω = I2ω.

From this, we can see that I1 = 2*I2.

The time taken to come to a complete stop is proportional to the moment of inertia, so the plate will take twice as long to stop when rotating about an axis through its center compared to when it is rotating about an axis through one of its corners.

So, if the time taken to stop when rotating about a corner is t1, the time taken to stop when rotating about the center will be 2*t1.