Question

A ball on a string moves around a complete circle, once a second, on a frictionless , horizontal table. The tension in the string is measured to be 6.0 N. What would the tension be if the ball went around in only half a second?

A 1.5 N
B 3.0 N
C 12 N
D 24 N

Answer 1

The tension in the string will be 24 N if the ball rotates twice as fast, completing a circle in half a second, due to the square relationship between tension and velocity.

Step-by-step explanation:

The tension in the string can be analyzed using the formula for centripetal force: F = m × v^2 / r, where F is the centripetal force (or tension), m is the mass of the ball, v is the tangential velocity, and r is the radius of the circle. Because only the period of the ball's rotation is changing and not its mass or the radius of the circle, we can forego the mass and radius values and focus on the relationship between tension and velocity.

When the ball moves around the circle in half the original time (0.5 seconds instead of 1 second), its velocity doubles. Since tension is proportional to the square of the velocity (v^2), if we double the velocity, the tension will increase by a factor of 2^2, which is 4. Therefore, if the original tension was 6.0 N, the new tension when the ball rotates twice as fast would be 4 times larger: 6.0 N × 4 = 24 N.