Question

Jannel swings a 175-g ball on the end of a light string causing it to revolve uniformly above her head in a horizontal circle of diameter 1.0 m. The ball makes 2.0 revolutions every 1.0 s. Find the tension in the string. (Hint: Find the acceleration needed to keep the ball in a circle).

Answer 1

1.4π² N

Step-by-step explanation:

Given:

m = 175 g = 0.175 kg

r = 1/2 × 1 = 0.5 m

ω = 2(2π) = 4π rad/s

`\boxed{\Sigma F_s=ma_s=m\omega^2r}`

`\Sigma F_s=T`

`T=m\omega^2r`

`=0.175*(4\pi)^2*0.5`

`=1.4\pi^2\ N`

Answer 2

To find the tension in the string of a ball in circular motion, calculate the acceleration needed to maintain its path. Determine the angular velocity, use it to calculate centripetal acceleration, and then multiply by the mass of the ball to find the tension.

Step-by-step explanation:

Finding the Tension in the Revolving Ball's String

To find the tension in the string of a ball revolving uniformly in a horizontal circle, we first need to find the acceleration necessary to keep the ball moving in a circle. Janell swings a 175-g ball (0.175 kg) in a horizontal circle with a diameter of 1.0 m, indicating a radius (r) of 0.5 m. The ball makes 2 revolutions per second, which translates to a frequency (f) of 2 Hz.

The first step is to calculate the angular velocity (ω) using the formula ω = 2πf. In this case, ω = 2π(2) = 4π rad/s. Next, the centripetal acceleration (ac) can be calculated using ac = ω2r. Plugging in our values gives us ac = (4π)2(0.5) m/s2.

Now we can determine the tension (T) in the string, which is the centripetal force required to keep the ball moving in a circle. The centripetal force is given by Fc = mac, where m is the mass of the ball and ac is the centripetal acceleration. Thus, T = Fc = m(4π)2(0.5).

Substituting the given mass of the ball, we find T = 0.175 kg ⋅ (4π)2 ⋅ 0.5 m/s2. After calculating, the resulting tension in the string is the force required to keep the ball in circular motion.