Question

In a track and field event, a hammer thrower accelerates the hammer (mass = 7.30 kg) from rest within four full turns (revolutions) and releases it at a speed of 22.0 m/s. Assuming a uniform rate of increase in angular velocity and a radius of 1.30 m, calculate the angular acceleration.

Answer 1

Answer:

5.69755 rad/s²

Explanation:

r = Radius = 1.3 m

v = Velocity of the hammer = 22 m/s

n = Number of revolutions = 4

Angular displacement

$\theta=n\times 2\pi\\\Rightarrow \theta=4\times 2\pi\\\Rightarrow \theta=8\pi$

$\omega_i$ = Initial angular speed = 0

Final angular speed

$\omega_f=\frac{v}{r}\\\Rightarrow \omega_f=\frac{22}{1.3}\\\Rightarrow \omega_f=16.92307\ rad/s$

Angular acceleration

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{16.92307^2-0^2}{2\times 8\pi}\\\Rightarrow \alpha=5.69755\ rad/s^2$

Angular acceleration is given by 5.69755 rad/s²