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

5.69755 rad/s²

Step-by-step explanation:

r = Radius = 1.3 m

v = Velocity of the hammer = 22 m/s

n = Number of revolutions = 4

Angular displacement

`\theta=n* 2\pi\\\Rightarrow \theta=4* 2\pi\\\Rightarrow \theta=8\pi`

`\omega_i` = Initial angular speed = 0

Final angular speed

`\omega_f=(v)/(r)\\\Rightarrow \omega_f=(22)/(1.3)\\\Rightarrow \omega_f=16.92307\ rad/s`

Angular acceleration

`\alpha=(\omega_f^2-\omega_i^2)/(2\theta)\\\Rightarrow \alpha=(16.92307^2-0^2)/(2* 8\pi)\\\Rightarrow \alpha=5.69755\ rad/s^2`

Angular acceleration is given by 5.69755 rad/s²