The iron ball shown is being swung in a vertical circle at the end of a 0.7-m string.How slowly can the ball go through its top position without having the string go slack?
A)1.3 m/s
B)2.6 m/s
C)3.9 m/s
D)6.9 m/s
E)9.8 m/s

Question

The iron ball shown is being swung in a vertical circle at the end of a 0.7-m string.How slowly can the ball go through its top position without having the string go slack? 
A)1.3 m/s 
B)2.6 m/s 
C)3.9 m/s 
D)6.9 m/s 
E)9.8 m/s

Quizplus · Accepted Answer

At the top of the vertical circle, the tension in the string must be equal to the weight of the ball. If the ball goes too slow, gravity will be greater than the tension in the string and the string will go slack.

To find the minimum speed, we can use the centripetal force equation:

F = mv^2/r

where F is the tension in the string, m is the mass of the ball, v is its velocity, and r is the radius of the circle (which is equal to the length of the string, 0.7 m).

At the top of the circle, we have:

F = mg

where g is the acceleration due to gravity (9.8 m/s^2).

We can substitute this into the centripetal force equation and solve for v:

mg = mv^2/r

v^2 = gr

v = sqrt(gr)

v = sqrt(9.8 m/s^2 * 0.7 m)

v ≈ 2.6 m/s

Therefore, the ball must go through its top position at a speed of at least 2.6 m/s to avoid having the string go slack.

The Iron Ball Shown Is Being Swung in a Vertical