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Question: A torque of 30 N-m is acting on a wheel of mass 5 kg and moment of inertia \(2\,kg-{{m}^{2}}\). If w...

A torque of 30 N-m is acting on a wheel of mass 5 kg and moment of inertia 2 kg−m22\,kg-{{m}^{2}}. If wheel starts rotating from rest, then its angular displacement in 10 seconds will be
A) 750 rad
B) 1500 rad
C) 3000 rad
D) 6000 rad

Explanation

Solution

There is much information given in the question, first write all the information properly and then it will become easy to get the answer that is asked in the question. There are different formulas for torque, angular velocity, and angular displacement, so all the formulas should be used properly to get the final answer, as the result of each formula will help to get the final answer.

Formula used:
τ=MOI×α ωf=ωi+αt θ=ωit+12αt2 \begin{aligned} & \tau =MOI\times \alpha \\\ & {{\omega }_{f}}={{\omega }_{i}}+\alpha t \\\ & \theta ={{\omega }_{i}}t+\dfrac{1}{2}\alpha {{t}^{2}} \\\ \end{aligned}

Complete solution:
According to the question,
Torque, τ=30 N−m Mass, m=5 kg Moment of Inertia, MOI=2 kg−m2 Time, t=10 seconds Initial Angular Velocity, ωi=0 \begin{aligned} & \text{Torque, }\tau =30\,N-m \\\ & \text{Mass,}\,m=5\,kg \\\ & \text{Moment of Inertia,}\,MOI=2\,kg-{{m}^{2}} \\\ & \text{Time, }t=10\,seconds \\\ & \text{Initial Angular Velocity, }{{\omega }_{i}}=0 \\\ \end{aligned}

Since, torque can be given as moment of inertia multiplied by angular acceleration, which is given as:
τ=MOI×α\tau =MOI\times \alpha

Substituting the value mentioned above, we get,
30 N−m=2 kg−m2×α ∴α=15 N/kg−m   \begin{aligned} & 30\,N-m=2\,kg-{{m}^{2}}\times \alpha \\\ & \therefore \alpha =15\,{N}/{kg-m}\; \\\ \end{aligned}
Now,

Angular velocity of the wheel after 10 seconds is,
ωf=ωi+αt{{\omega }_{f}}={{\omega }_{i}}+\alpha t

Substituting the value mentioned above, we get –
ωf=(0+15×10) rad/sec   ∴ωf=150 rad/sec   \begin{aligned} & {{\omega }_{f}}=(0+15\times 10)\,{rad}/{sec}\; \\\ & \therefore {{\omega }_{f}}=150\,{rad}/{sec}\; \\\ \end{aligned}

And Angular displacement of the wheel after 10 seconds is:
θ=ωit+12αt2\theta ={{\omega }_{i}}t+\dfrac{1}{2}\alpha {{t}^{2}}

Substituting the value mentioned above, we get –
θ=(0×10+12×15×102) rad ⇒θ=(0+12×15×100) rad ∴θ=750 rad \begin{aligned} & \theta =(0\times 10+\dfrac{1}{2}\times 15\times {{10}^{2}})\,rad \\\ & \Rightarrow \theta =(0+\dfrac{1}{2}\times 15\times 100)\,rad \\\ & \therefore \theta =750\,rad \\\ \end{aligned}

Therefore, the correct answer is Option (A).

Note:
There are many questions in which different types of information are given and students get confused after seeing so much details on which formula to apply to get the final answer, so it’s better to list down all the details given in the question properly first and then choose which formula should be used to get the desired result. Combination of formula can also be used to achieve the result that is asked in the question, as the outcome of one formula can be used to get the final outcome.