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Rotational Motion question

2022 · 28 Jun · Shift 2 · Q51
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  5. /2022 · 28 Jun · Shift 2 · Q51

Rotational Motion question

2022 · 28 Jun · Shift 2 · Q51

JEE MainPhysicsRotational MotionMCQ+4 / −1
A ball is spun with angular acceleration α\alphaα= 6t2 −-− 2t where t is in second and α\alphaα is in rads −-− 2. At t = 0, the ball has angular velocity of 10 rads −-− 1 and angular position of 4 rad. The most appropriate expression for the angular position of the ball is :
  1. A
    32t4−t2+10t{3 \over 2}{t^4} - {t^2} + 10t23​t4−t2+10t
  2. B
    t42−t33+10t+4{{{t^4}} \over 2} - {{{t^3}} \over 3} + 10t + 42t4​−3t3​+10t+4
  3. C
    2t43−t36+10t+12{{2{t^4}} \over 3} - {{{t^3}} \over 6} + 10t + 1232t4​−6t3​+10t+12
  4. D
    2t4−t32+5t+42{t^4} - {{{t^3}} \over 2} + 5t + 42t4−2t3​+5t+4
View written solutionFree

Correct answer: B

  1. Given data
  • Angular acceleration: α(t)=6t2−2t\alpha(t)=6t^2-2tα(t)=6t2−2t
  • Initial angular velocity: ω(0)=10 rad s−1\omega(0)=10\ \text{rad s}^{-1}ω(0)=10 rad s−1
  • Initial angular position: θ(0)=4 rad\theta(0)=4\ \text{rad}θ(0)=4 rad

We need to find the angular position θ(t)\theta(t)θ(t).

  1. Find angular velocity by integrating angular acceleration

We know: α=dωdt\alpha=\frac{d\omega}{dt}α=dtdω​ So, ω(t)=∫(6t2−2t) dt\omega(t)=\int (6t^2-2t)\,dtω(t)=∫(6t2−2t)dt

Integrating, ω(t)=2t3−t2+C1\omega(t)=2t^3-t^2+C_1ω(t)=2t3−t2+C1​

Now use the initial condition ω(0)=10\omega(0)=10ω(0)=10: 10=2(0)3−(0)2+C110=2(0)^3-(0)^2+C_110=2(0)3−(0)2+C1​ C1=10C_1=10C1​=10

Thus, ω(t)=2t3−t2+10\omega(t)=2t^3-t^2+10ω(t)=2t3−t2+10

  1. Find angular position by integrating angular velocity

We know: ω=dθdt\omega=\frac{d\theta}{dt}ω=dtdθ​ So, θ(t)=∫(2t3−t2+10) dt\theta(t)=\int (2t^3-t^2+10)\,dtθ(t)=∫(2t3−t2+10)dt

Integrating, θ(t)=2t44−t33+10t+C2\theta(t)=\frac{2t^4}{4}-\frac{t^3}{3}+10t+C_2θ(t)=42t4​−3t3​+10t+C2​

Simplify: θ(t)=t42−t33+10t+C2\theta(t)=\frac{t^4}{2}-\frac{t^3}{3}+10t+C_2θ(t)=2t4​−3t3​+10t+C2​

Now use the initial condition θ(0)=4\theta(0)=4θ(0)=4: 4=0−0+0+C24=0-0+0+C_24=0−0+0+C2​ C2=4C_2=4C2​=4

Hence, θ(t)=t42−t33+10t+4\boxed{\theta(t)=\frac{t^4}{2}-\frac{t^3}{3}+10t+4}θ(t)=2t4​−3t3​+10t+4​

  1. Match with the options

Option B is: t42−t33+10t+4\frac{t^4}{2}-\frac{t^3}{3}+10t+42t4​−3t3​+10t+4

This matches exactly.

Therefore, the correct option is: B\boxed{\text{B}}B​

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