# Thread: Angular Velocity

1. ## Angular Velocity

Need help on this please folks, it has got me stumped.

1. 1. A body of mass m kg is attached to a point by string of length 1.25m. If the mass is rotating in a horizontal circle 0.75m below the point of attachement, calculate its angular velocity.

2. If the mass rotates on a table, calculate the force on the table when the speed of rotation is 25rpm and the mass is 6kg.

2. Originally Posted by DanBrown100
1. 1. A body of mass m kg is attached to a point by string of length 1.25m. If the mass is rotating in a horizontal circle 0.75m below the point of attachement, calculate its angular velocity.
you should be familiar with the forces acting on the "bob" (mass) of a conical pendulum ...

if $\displaystyle L = 1.25 \, m$ , $\displaystyle h = 0.75 \, m$, then $\displaystyle r = 1.0 \, m$

$\displaystyle T\cos{\theta} = mg$

$\displaystyle T\sin{\theta} = mr \omega^2$

you now have enough information to solve for $\displaystyle \omega$.

2. If the mass rotates on a table, calculate the force on the table when the speed of rotation is 25rpm and the mass is 6kg.
this part is not clear ... how are the string and pendulum bob oriented?

3. ## Re: Angular Velocity

Originally Posted by DanBrown100
Need help on this please folks, it has got me stumped.

1. 1. A body of mass m kg is attached to a point by string of length 1.25m. If the mass is rotating in a horizontal circle 0.75m below the point of attachment, calculate its angular velocity.

2. If the mass rotates on a table, calculate the force on the table when the speed of rotation is 25rpm and the mass is 6kg.
Part 2.

I assume for the second part that the radius of the circle is the same as in Part 1. The table supplies the extra force needed since the tension T will be reduced. The force the ball exerts on the table is equal in magnitude to the force the table exerts on the ball (but in the opposite direction).

Following what skeeter gave, you then have:

$\displaystyle T_2\sin{\theta} = m\,r\,{\omega_2}^2$

$\displaystyle T_2\cos{\theta} +F_T = mg$

$\displaystyle \omega_2=60\cdot 2\pi\cdot 25$  in units of radians per second.