Question

A pulley of radius 1.5 m is rotated about its axis by a force F = (12t – 3t²) N applied tangentially (while t is measured in seconds). If moment of inertia of the pulley about its axis of rotation is 4.5 kg m², the number of rotations made by the pulley before its direction of motion is reversed, will be K/π. The value of K is ________.

Answer 1

Correct Answer: 18

To solve the problem, we first determine the angular acceleration α of the pulley using the relationship between torque τ and moment of inertia — τ = Iα. The tangential force F is given as a function of time t: F = (12t – 3t²) N. Since torque τ is related to the force and radius r by τ = Fr, we have:  
τ = (12t - 3t2) × 1.5 = 18t - 4.5t2 Nm.
Given that I = 4.5 kg m², the angular acceleration is computed as:
α = τ/I = (18t - 4.5t2)/4.5 = 4t - t2 rad/s2.
To find when the direction reverses, set the angular velocity ω to zero and integrate the expression for α: 
ω = ∫(4t - t2)dt = 2t2 - (t3/3).
The initial angular velocity is zero, and we solve for t when ω again becomes zero:
2t2 - (t3/3) = 0.
Factor the equation:
t2(2 - t/3) = 0.
Thus, solutions are t=0 or t=6. For t=0, it’s the initial condition. We use t=6. Calculate the angular displacement θ by integrating:
θ = ∫ω dt = ∫(2t2 - t3/3) dt = (2t3/3) - (t4/12) |60.
Evaluate the definite integral:
θ = [(2×63/3) - (64/12)] - [0] = (2×216/3) - (1296/12).
This simplifies to:
θ = 144 - 108 = 36 radians.
The number of full rotations N is:
N = θ/2π = 36/2π = 18/π.
This gives N = K/π, and thus K = 18. The computed value falls within the provided range of 18,18, confirming the correctness of the value.

Answer 2

The correct answer is 18

Fig.

_{I =} 4.5 kg m²
\(FR=Iα\)
\(α=\frac{(12t–3t^2)×1.5}{4.5}=4t−t^2\)
\(w=∫αdt=2t^2–\frac{t^3}{3} \)
w=0
\(⇒t^2(2–\frac{t}{3})0 \)
t=6 sec
\(θ=∫_{0}^{6}[2^t2–\frac{t^3}{3}]dt=[\frac{2t^3}{3}–\frac{t^4}{12}]_{0}^{6}\)
\(=[\frac{2}{3}×6^3–\frac{6^4}{12}]=36\)
\(n=\frac{36}{2π}\)
\(=\frac{18}{π}\)