Question

A mass m rotating freely in a horizontal circle of radius 1m on a friction less smooth table supports a stationary mass 2m attached to the other end of the string passing through smooth hole O in the table hanging vertically, find the angular velocity of rotation

Answer 1

Answer:

Step 1:

Mass of the stationary object is = 2 m

Here, there are only two forces acting on to the body, the downward gravitational force of “mg” and the tension in the string that holds the mass up.

∴ T = 2mg …… (i)

Step 2:

The mass of the rotating object is = m

The radius of the circle, r = 1 m

The force acting on the mass will be the same tension in the string as it is the same string connecting the two masses through a hole. So, the centripetal force is equal to the tension for the rotating mass on the table.

Therefore,  

The centripetal force or the tension for the rotating mass on the table,

T = mv² / r …. (ii)

Step 3:

Equating eq. (i) & (ii), we get

2mg = mv²/r

Substituting the values

⇒ v = √[2gr] = √[2*9.8*1] = √[19.6]

Step 4:

Thus,  

The angular velocity, ω = v/r = √[19.6] / 1 = 4.42 rad/sec