a A body of mass m kg is attached to a point by string of le

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

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

Solution

L = 1.25m
h = 0.75m
= ?
-------------------
String is positioned at angle to vertical,
cos = h/L = 0.75/1.25 = 0.6
we may calculate angle , but it\'s not necessary, as later we\'ll need cos, not .

Rotating mass is exposed to three forces:
- centrifugal force Fcf = m²r (where r is radius of rotation)
- weight G=mg
- tension in string, S
In state of equilibrium of forces, the sum of torques of these forces around point of attachment is zero:
Fcf L cos - G L sin = 0
(force S does not produce torque to point of attachment because its direction passes through that point)
Fcf cos = G sin
m ² r cos = m g sin
masses cancel each other, and we put r = L sin
² L sin cos = g sin
sin cancels
² = g / (L cos )
= [g / (L cos )]
= [9.81 / (1.25 * 0.6)] = 3.617 rad/s

b).