Question

A 0.100 kg ball is swung in a vertical circle on the end of its 0.50 m long string.

a. where in the rotation will the string experience the most tension, and why?

b. At that point, how would you determin the amount of tension in the string given the frequency of rotaion?

c. if the string can apply up to 50.0 N in tension, state the maximum speed of rotation of the ball that the string can withstand

Answer 1

a. The string will experience the most tension at the top of the ball's path because the tension in the string must provide the centripetal force required to keep the ball moving in a circle against the weight of the ball.

b. At the top of the ball's path, the tension in the string is equal to the weight of the ball, which is mg.

c. The maximum speed of rotation that the ball can have without exceeding the maximum tension in the string, which is 50.0 N, is 7.0 m/s.

Step-by-step explanation:

a. The tension in the string will be greatest at the top of the ball's path. This is because at the top of the path, the weight of the ball is acting downwards, and the tension in the string must be large enough to provide the centripetal force required to keep the ball moving in a circle. At the bottom of the path, the tension in the string is less because the weight of the ball is aiding the centripetal force.

b. At the top of the ball's path, the tension in the string can be determined using the following equation:

⇒ T = m(g + v^2/r)

where T is the tension in the string, m is the mass of the ball, g is the acceleration due to gravity, v is the speed of the ball, and r is the radius of the circle (in this case, the length of the string). At the top of the path, the speed of the ball is momentarily zero, so the tension in the string is equal to:

⇒ T = m(g + 0/r) = mg

c. To find the maximum speed of rotation that the string can withstand, we can use the same equation as in part (b), but solve for v:

v = sqrt((T - mg)r/m)

where T is the maximum tension the string can withstand, m is the mass of the ball, g is the acceleration due to gravity, and r is the length of the string. Substituting the given values, we get:

v = sqrt((50 N - (0.100 kg)(9.81 m/s^2))(0.50 m)/(0.100 kg)) = 7.0 m/s

Therefore, the maximum speed of rotation that the ball can have without exceeding the maximum tension in the string is 7.0 m/s.

Answer 2

a. The tension in the string will be greatest at the lowest point of the circle. This is because at this point, the weight of the ball is pulling directly downwards, and the tension in the string must be great enough to provide the centripetal force required to keep the ball moving in a circle.

b. At the lowest point in the circle, the tension in the string can be found using the equation:

T = mv^2/r + mg

where T is the tension in the string, m is the mass of the ball, v is the speed of the ball, r is the radius of the circle (equal to the length of the string), and g is the acceleration due to gravity.

At the lowest point in the circle, the speed of the ball can be found using the equation for centripetal acceleration:

a = v^2/r

where a is the centripetal acceleration.

Combining these two equations, we get:

T = ma + mg

T = m(v^2/r) + mg

c. To find the maximum speed of rotation that the string can withstand, we can rearrange the equation from part (b) to solve for v:

v = sqrt((T - mg)r/m)

Substituting the given values, we get:

v = sqrt((50.0 N - 0.100 kg x 9.81 m/s^2) x 0.50 m / 0.100 kg)

v = sqrt(24.5)

v ≈ 4.95 m/s

Therefore, the maximum speed of rotation that the string can withstand is approximately 4.95 m/s.