A cylinder having a mass of 3 kg can rotate about its central axis through point

Question

A cylinder having a mass of 3 kg can rotate about its central axis through point O. Forces are applied as shown figure. F1= 3N   F2=2N.   F3= 1N   F4= 2N
Also R1= 5 cm and R2 = 12cm . Find the magnitude of acceleration of the cylindet

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Explanation / Answer

There is an equation similar to "F=ma" that deals with angular motion. But instead of force (F), it uses torque (); instead of mass (m), it uses "moment of inertia" (I); and instead of acceleration (a), it uses "angular acceleration" ().

Torque = (moment of inertia) × (angular acceleration)

= I

Torque equals (force)×(lever arm). The "lever arm" is the perpendicular separation between the line of force and the axis.

In the diagram, there are four forces, therefore four torques. You need to add them all up to get the total torque ()

Torque_1 = F1 × R2 (positive because it pulls counter-clockwise). = 36
Torque_2 = F2 × R2 (negative because it pulls clockwise) = -24
Torque_3 = F3 × r (negative because it pulls clockwise). = -6
Torque_4 = F4 × 0 (zero because the lever arm is zero). = 0

Add up those four torques to get "". = 6N-cm = 0.06N-m

Next, you need to know what the moment of inertia of the cylinder is. From your book, you can find that the moment of inertia of a solid cylinder is:

I = mR²/2 = 216kg-cm^2 = 2.16kg-m^2

(where "m" is the cylinder's mass).

= I

= 36 rad/s^2

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