Physics · System Of Particles And Rotational Motion · NEET
Angular acceleration alpha is always in rad/s^2, never m/s^2. m/s^2 is the unit of linear (tangential) acceleration a. NTA often puts a wrong m/s^2 option to trap you. The formula alpha = FR/I gives units of (N.m)/(kg.m^2) = 1/s^2 = rad/s^2. So pick the rad/s^2 option. If a question asks for the linear acceleration of a point on the rim, then use a = R.alpha, which comes in m/s^2.
Read the wording carefully. A hollow cylinder (or thin ring/hoop about its axis) has all mass at radius R, so I = MR^2. A solid cylinder (or disc about its axis) has I = (1/2)MR^2. Using the wrong one changes alpha by a factor of 2. In the 2017 NEET question the cylinder is hollow, so I = MR^2, giving alpha = 25 rad/s^2. If it were solid, the same numbers would give 50 rad/s^2.
The rope leaves the cylinder tangentially, so the pulling force F is tangent to the rim. The position vector from the axis to the rim point is along the radius R, which is perpendicular to a tangent. So the angle between R and F is 90 degrees, and sin(90) = 1. That is why torque = F.R.sin(90) = FR. The rim is the smart place to apply force because it gives the maximum torque for that force.
No. In NEET problems the rope is taken as massless (light) and it does not slip on the cylinder. So the rope only transmits the force F to the rim of the cylinder. You only use the cylinder's mass M in I = MR^2 (hollow) or (1/2)MR^2 (solid). If the rope had a hanging block, then you would set up two equations (block: mg - T = ma, cylinder: TR = I.alpha) and use a = R.alpha to connect them.
When you pull the free end of a light rope directly with force F (no hanging mass, no friction losses), the tension throughout the rope equals F, and this full F acts tangentially on the rim. So torque = FR. But if the rope hangs over the cylinder with a block on the other end, the tension T is less than the block's weight, and you must solve for T first. Always check whether the number given is a direct pull force or a hanging weight.
A rope wound on a hollow cylinder of mass 3 kg and radius 40 cm is pulled with a force of 30 N. The angular acceleration of the cylinder is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
alpha = FR / I, where F is the pulling force, R is the cylinder radius, and I is the moment of inertia. For a hollow cylinder I = MR^2, so alpha = FR/(MR^2) = F/(MR). For a solid cylinder I = (1/2)MR^2, so alpha = 2F/(MR).
The rope is wound around the outer surface, so it leaves the cylinder at radius R. The pull is transmitted to that rim point, and it acts tangent to the circle, giving torque = FR.
Radians per second squared (rad/s^2). Do not confuse it with linear acceleration a in m/s^2. They are linked by a = R.alpha for a point on the rim.
The rope point on the rim has tangential acceleration a = R.alpha. Using the NEET 2017 numbers, a = 0.4 x 25 = 10 m/s^2. This is the rate at which the rope speeds up as it comes off the cylinder.
A solid cylinder has half the moment of inertia, I = (1/2)MR^2. With the same F and R, alpha doubles. So the NEET 2017 setup would give alpha = 50 rad/s^2 if the cylinder were solid.
