Question

A ball is spun with angular acceleration α = 6t² – 2t, where t is in second and α 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 :

• A. \(\frac{3}{4}t^4−t^2+10t\)
• B. \(\frac{t_4}{2}−\frac{t_3}{3}+10t+4\)
• C. \(\frac{2t^4}{3}−\frac{t^3}{6}+10t+12\)
• D. \(2t^4−\frac{t^3}{2}+5t+4\)

Answer 1

The Correct Option is B

To find the expression for the angular position of the ball, we need to integrate the given angular acceleration to obtain angular velocity and then integrate again to get the angular position. Let's go through the steps one by one:

1. Given: Angular acceleration \(\alpha = 6t^2 - 2t\). 
2. Integration to find angular velocity:

Integrate the angular acceleration to find velocity: \[ \omega(t) = \int \alpha \, dt = \int (6t^2 - 2t) \, dt = 2t^3 - t^2 + C_1 \] Given that at \(t = 0\), \(\omega(0) = 10 \text{ rads}^{-1}\), we find \(C_1\): \[ 10 = 2(0)^3 - (0)^2 + C_1 \implies C_1 = 10 \] Thus, \[ \omega(t) = 2t^3 - t^2 + 10 \]

1. Integration to find angular position:

Integrate the angular velocity to find the position: \[ \theta(t) = \int \omega(t) \, dt = \int (2t^3 - t^2 + 10) \, dt = \frac{2t^4}{4} - \frac{t^3}{3} + 10t + C_2 \] Simplifying, we get: \[ \theta(t) = \frac{t^4}{2} - \frac{t^3}{3} + 10t + C_2 \] Given that at \(t = 0\), \(\theta(0) = 4 \text{ rad}\), we find \(C_2\): \[ 4 = \frac{0^4}{2} - \frac{0^3}{3} + 10 \times 0 + C_2 \implies C_2 = 4 \] Thus, \[ \theta(t) = \frac{t^4}{2} - \frac{t^3}{3} + 10t + 4 \]

Therefore, the expression for the angular position of the ball is \(\frac{t^4}{2} - \frac{t^3}{3} + 10t + 4\).

The correct answer is: \(\frac{t^4}{2} - \frac{t^3}{3} + 10t + 4\).

Answer 2

The correct answer is (B) : 

\[\frac{t_4}{2} - \frac{t_3}{3} + 10t + 4\]

\[ \alpha = \frac{d\omega}{dt} = 6t^2 - 2t \]

\[ \int_{0}^{\omega} d\omega = \int_{0}^{t}(6t^2 - 2t) dt \]

so \( \omega = 2t^3 - t^2 + 10 \)

and

\[ \frac{d\theta}{dt} = 2t^3 - t^2 + 10 \] 

so

\[ \int_{4}^{\theta} d\theta = \int_{0}^{t} (2t^3 - t^2 + 10) dt \]

\[ \theta = \frac{t_4}{2} - \frac{t^3}{3} + 10t + 4 \]