Question

A wheel of diameter 40 cm starts from rest and attains a speed of 240 rpm in 4 mins. Calculate its angular displacement within this time interval.

Answer 1

Answer: $\theta = 3014.4\ rad$

Explanation:

Given that,

Initial angular velocity

$\omega_{0} = 0$

Final angular velocity

$\omega_{f} = \dfrac{2\pi N}{60}$

$\omega_{f} = \dfrac{2\times3.14\times 240\ rpm}{60\ s}$

$\omega_{f} = 25.12\ rad/s$

The average angular velocity

$\omega = \dfrac{\omega _{i}+\omega_{f}}{2}$

$\omega = 12.56\ rad/s$

Now, the angular displacement is

$\theta = \omega t$

$\theta = 12.56\ rad/s\times4\times60\ s$

$\theta = 3014.4\ rad$

Hence, the angular displacement is $\theta = 3014.4\ rad$

Answer 2

Hey dear,

● Answer -
θ = 3016 rad

● Explaination -
# Given -
r = 20 cm = 0.2 m
f1 = 0
f2 = 240 rpm = 4 Hz
t = 4 min = 240 s

# Solution -
Average angular velocity is calculated by -
ω = (ω1 + ω2) / 2
ω = 2π(f1 + f2) / 2
ω = 2 × 3.14 (0 + 4) / 2
ω = 12.56 rad/s

Angular displacement in 4 min -
θ = ωt
θ = 12.56 × 240
θ = 3016 rad

Therefore, angular displacement in 4 min is 3016 rad.

Hope this helps you...