Physics · System Of Particles And Rotational Motion · NEET
Multiply rpm by 2*pi to get radians per minute, then divide by 60 to get radians per second. So omega0 = rpm * 2*pi / 60. Example: 3 rpm = 3 * 2*pi / 60 = 0.1*pi rad/s. Never plug rpm straight into a physics formula, because every rotational equation uses SI units (rad/s). This one conversion is where most NEET students lose the mark.
Because the question gives you the number of revolutions (an angle theta), not the time t. Pick the kinematic equation that contains the quantities you are given and the one you want. Here you know omega0, final omega (=0 for stopping), and theta, and you want alpha, so omega^2 = omega0^2 - 2*alpha*theta is the direct route. If time were given instead, you would use omega = omega0 - alpha*t.
Always radians. One revolution = 2*pi radians, so N revolutions = N * 2*pi radians. If a body stops in 2 revolutions, theta = 2 * 2*pi = 4*pi rad. Leaving theta as '2' is a classic silent error, the number looks reasonable but is wrong by a factor of 2*pi.
Use the I of the given shape about the given axis. For a solid cylinder or disc about its central axis, I = (1/2)*M*R^2. For a ring or hollow cylinder, I = M*R^2. For a solid sphere about a diameter, I = (2/5)*M*R^2. Convert the radius to metres first (4 cm = 0.04 m), or your I will be off by a large power of ten.
The retardation alpha is opposite to the spin, so technically the torque is negative (it opposes motion). But NEET asks for the magnitude of the required torque, so report the positive value tau = I * alpha. Just remember the torque acts against the rotation, that is what brings the body to rest.
A solid cylinder of mass 2 kg and radius 4 cm rotates about its axis at 3 rpm. The torque required to stop it within 2 revolutions is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
tau = I * alpha, where I is the moment of inertia and alpha is the angular retardation found from omega^2 = omega0^2 - 2*alpha*theta with final omega set to zero.
N revolutions = N * 2*pi radians. So 1 rev = 2*pi rad, 2 rev = 4*pi rad, and so on. Always convert revolutions to radians before using any rotational equation.
I = (1/2)*M*R^2. This is the same as a solid disc about its central axis, since a cylinder is just a thick disc.
Yes, indirectly through the moment of inertia. Larger mass or larger radius means larger I, so more torque is needed to stop it in the same number of revolutions.
It combines three ideas in one question: unit conversion (rpm to rad/s), rotational kinematics (omega^2 = omega0^2 - 2*alpha*theta), and rotational dynamics (tau = I*alpha). NEET repeatedly tests this chain, as in the 2019 solid-cylinder problem.
