Question

A uniform rod of mass M = 1.2 kg and length L = 0.80 m, lying on a frictionless horizontal plane, is free to pivot about a vertical axis through one end, as shown. The moment of inertia of the rod about this axis is given by (1/3) ML 2. If a force ( F = 5.0 N, θ = 40 °) acts as shown, what is the resulting angular acceleration about the pivot point?

Answer 1

We have that for the Question it can be said that  the resulting angular acceleration about the pivot point is

• 10.04 rad/s^2  

Option D

From the question we are told

A uniform rod of mass M = 1.2 kg and length L = 0.80 m, lying on a frictionless horizontal plane, is free to pivot about a vertical axis through one end, as shown. The moment of inertia of the rod about this axis is given by (1/3) ML 2. If a force ( F = 5.0 N, θ = 40 °) acts as shown, what is the resulting angular acceleration about the pivot point?

Generally the equation for the moment of inertia  is mathematically given as

$I =1/3* ML^2\\\\Therefore\\\\I=(1.2*0.8^2)/3\\\\I=0.256$

Therefore

$\omega=\frac{ Lfsin(\theta)}{I}\\\\Therefore\\\\\omega=\frac{ 0.8*5*sin(40)}{0.256}\\\\\omega=10.04rads/s^2$

Option D

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Answer 2

The resulting angular acceleration about the pivot point is 10.04 rad/s².

The given parameters;

• mass of the rod, m = 1.2 kg
• length of the rod, L = 0.8 m

The moment of inertia of the uniform rod is calculated as follows;

$I = \frac{1}{3} ML^2\\\\I = \frac{1}{3} \times 1.2 \times (0.8)^2\\\\I = 0.256 \ kgm^2$

The angular acceleration is determine by applying formula for calculating torque;

$\tau = I \alpha \\\\F.Lsin(\theta) = I \alpha \\\\\alpha \ = \frac{F.Lsin(\theta)}{I} \\\\\alpha \ = \frac{5 \times 0.8\times sin(40)}{0.256} \\\\\alpha \ = 10.04 \ rad/s^2$

Thus, the resulting angular acceleration about the pivot point is 10.04 rad/s².

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