A pulley of radius 2m is rotated about its axis by a force F = (20t – 5t2)
newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2, the number of rotations made by the pulley before its direction of motion if reversed, is :
To reverse the direction òtdq = 0 (work done is zero)
t = (20 t – 5t2) 2 = 40t – 10t2
\(\alpha = \frac{\tau }{I} = \frac{{40t - 10{t^2}}}{{10}} = 4t - {t^2}\)
\(\omega = \int_0^t {\alpha dt = 2{t^2} - \frac{{{t^3}}}{3}} \)
w is zero at
\(2{t^2} - \frac{{{t^3}}}{3} = 0\)
t3 = 6t2
t = 6 sec.
q = ò wdt
\( = \int_0^6 {\left( {2{t^2} - \frac{{{t^3}}}{3}} \right)dt} \)
\(\left[ {\frac{{2{t^3}}}{3} - \frac{{{t^4}}}{{12}}} \right]_0^6 = 216\left[ {\frac{2}{3} - \frac{1}{3}} \right] = 36\,rad.\)
No or revolution \(\frac{{36}}{{2\pi }}\) Less than 6
This problem involves rotational motion where a tangential force causes angular acceleration, and we need to find the number of rotations before the pulley reverses direction. The reversal occurs when the angular velocity becomes zero after being positive.
The torque τ acting on the pulley is given by τ = F × r, where F is the tangential force and r is the radius.
Given: F = (20t – 5t²) N, r = 2 m
So, τ = (20t – 5t²) × 2 = (40t – 10t²) N·m
From Newton's second law for rotation: τ = Iα, where I is the moment of inertia and α is the angular acceleration.
Given: I = 10 kg·m²
Therefore, α = τ / I = (40t – 10t²) / 10 = (4t – t²) rad/s²
Angular acceleration α is the derivative of angular velocity ω with respect to time: α = dω/dt
So, dω/dt = 4t – t²
Integrating both sides with respect to time t:
ω = ∫(4t – t²) dt = 2t² – (t³/3) + C
Assuming the pulley starts from rest, at t=0, ω=0. So, C=0.
Thus, ω = 2t² – (t³/3)
The direction reverses when angular velocity ω becomes zero.
Set ω = 0: 2t² – (t³/3) = 0
t²(2 – t/3) = 0
So, t=0 (initial time) or t=6 s (when reversal happens)
Angular displacement θ is the integral of ω from t=0 to t=6 s.
θ = ∫₀⁶ ω dt = ∫₀⁶ (2t² – t³/3) dt
Compute the integral:
∫(2t²) dt = (2/3)t³
∫(t³/3) dt = (1/12)t⁴
So, θ = [ (2/3)t³ – (1/12)t⁴ ] from 0 to 6
At t=6: (2/3)(216) – (1/12)(1296) = 144 – 108 = 36 rad
At t=0: 0
So, total angular displacement θ = 36 radians
Number of rotations N = θ / (2π) = 36 / (2π) = 18/π ≈ 18/3.1416 ≈ 5.73 rotations
Since 5.73 is more than 3 but less than 6, the correct option is "more than 3 but less than 6".
Torque in Rotation: τ = F × r (if force is tangential)
Newton's Second Law for Rotation: τ = Iα
Angular Kinematics: α = dω/dt, ω = dθ/dt
Integration to find ω and θ: ω = ∫α dt, θ = ∫ω dt
Conversion: Number of rotations = θ / (2π)