Question

1. a ball is attached to a string of length 1.5 m and spun in a horizontal circle. if it makes 120 turns per minute (i.e. 120 rev/min), find: the linear speed of the ball? its centripetal acceleration? t = .5

Explanation:

Step1: Find the angular velocity (ω)

First, convert the rotational speed from rev/min to rad/s. The frequency \( f = 120 \, \text{rev/min} = \frac{120}{60} = 2 \, \text{rev/s} \). The angular velocity \( \omega = 2\pi f \), so \( \omega = 2\pi \times 2 = 4\pi \, \text{rad/s} \).

Step2: Calculate the linear speed (v)

The formula for linear speed in circular motion is \( v = r\omega \), where \( r = 1.5 \, \text{m} \) (length of the string, which is the radius of the circle). Substituting the values, \( v = 1.5 \times 4\pi = 6\pi \approx 18.85 \, \text{m/s} \).

Step3: Calculate the centripetal acceleration (aₙ)

The formula for centripetal acceleration is \( a_n = \frac{v^2}{r} \) or \( a_n = r\omega^2 \). Using \( a_n = r\omega^2 \), substitute \( r = 1.5 \, \text{m} \) and \( \omega = 4\pi \, \text{rad/s} \). So \( a_n = 1.5 \times (4\pi)^2 = 1.5 \times 16\pi^2 = 24\pi^2 \approx 236.87 \, \text{m/s}^2 \).

Answer:

• Linear speed of the ball: \( \approx 18.85 \, \text{m/s} \) (or \( 6\pi \, \text{m/s} \))
• Centripetal acceleration: \( \approx 236.87 \, \text{m/s}^2 \) (or \( 24\pi^2 \, \text{m/s}^2 \))