Question

A 500 g ball swings in a vertical circle at the end of a 1.5-m-long string. When the ball is at the bottom of the circle, the tension in the string is 10 N. What is the speed of the ball at that point?

Answer 1

3.91m/s

Step-by-step explanation:

The mass of the ball(m)=500g = 0.5kg

The radius of the string is(r)=1.5m

The tension in the string is(T)=10N

The acceleration due to gravity =
`9.8m/s^(2)`

The tension in the string when the body is at the bottom is given by

`T = (mv^(2) )/(r)+mg`

To find the speed of the ball, we make v the subject of the formula

Therefore,
`v=\sqrt(r(T-mg))/(m)`

`v=\sqrt(1.5(10-0.5*9.8))/(0.5)`

`v=\sqrt(7.65)/(0.5)`

`v=\sqrt15.3=3.9115` ≈ 3.91m/s

Therefore, the speed of the ball is 3.91m/s

Answer 2

3.9096m/s

Step-by-step explanation:

Since the ball moved in circular path, then it will experience centripetal acceleration which can be expressed below

a= v^2/r.......eqn(1)

From second law of Newton, we know that the ball will experience both force of tension and that of gravity.

F= (T - Fg)...........eqn(2)

r= distance= 1.5m

Where T= force of tension= 10N

Fg= force of gravity

But we know that

F= ma...............eqn(3)

Where a= acceleration

m= mass of the object= 500g= 0.5kg

Substitute eqn(1) and (2) into 3

(T - Fg)= mv^2/r

v^2= (T - Fg)r/m

v^2= √(T - Fg)r/m

But Fg= mg

v^2= √[10 -( 0.5×9.81)])1.5/0.5

v=3.9096m/s

Hence, the speed of the ball at that point is 3.9096m/s