Question

a 1 kg ball is tied to the end of a 2 m long string. the ball is swung in a horizontal circle at an angular speed of 4 rad/s. what is the torque on the ball? a. 4 nm b. 8 nm c. 16 nm d. 32 nm e. 0 nm

Explanation:

Step1: Recall the formula for torque

Torque $\tau = r\times F_{\perp}$. For a ball moving in a horizontal circle with constant angular speed (uniform circular motion), the centripetal force $F_{c}=m\omega^{2}r$ acts towards the center. The angle $\theta$ between the position vector $r$ (from the center of the circle to the ball) and the centripetal force $F_{c}$ is $\theta = 180^{\circ}$ (or $\pi$ radians). The formula for the cross - product $r\times F = rF\sin\theta$.

Step2: Calculate the sine of the angle

Since $\theta = 180^{\circ}$, $\sin\theta=\sin(180^{\circ}) = 0$.

Step3: Calculate the torque

Using $\tau=rF\sin\theta$, and $F = m\omega^{2}r$ (centripetal force), we substitute $\sin\theta = 0$. So, $\tau=r(m\omega^{2}r)\times0=0$.

Answer:

E. \(0\space Nm\)