Question

A baseball of mass m = 0.41 kg is spun vertically on a massless string of length l = 0.87 m. the string can only support a tension of tmax = 7.5 n before it will break. what is the maximum possible speed of the ball at the bottom of the loop, in meters per second?

Answer 1

To find the maximum possible speed of the ball at the bottom of the loop, we need to consider the tension in the string. The tension in the string provides the centripetal force that keeps the ball moving in a circular path.

Step-by-step explanation:

To find the maximum possible speed of the ball at the bottom of the loop, we need to consider the tension in the string. The tension in the string provides the centripetal force that keeps the ball moving in a circular path. At the bottom of the loop, the tension in the string is at its maximum. Therefore, we can equate the tension to the centripetal force:

T = mv^2 / r

where T is the tension, m is the mass of the ball, v is the speed, and r is the radius of the circular path. In this case, the radius is equal to the length of the string, l.

Substituting the given values:

Tmax = (0.41 kg)v^2 / 0.87 m

Solving for v:

v^2 = (Tmax * 0.87 m) / 0.41 kg

Now, we can solve for v:

v = sqrt[(Tmax * 0.87 m) / 0.41 kg]

Plugging in the given value for Tmax:

v = sqrt[(7.5 N * 0.87 m) / 0.41 kg]

Simplifying the equation further gives the maximum possible speed of the ball at the bottom of the loop.