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. $\frac{3}{4}$t4-t2+10t
• B. $\frac{t^4}{2}$-$\frac{t^3}{3}$+10t+4
• C. $\frac{2t^4}{3}$-$\frac{t^3}{6}$+10t+12
• D. 2t4-$\frac{t^3}{2}$+5t+4

Answer 1

Answer: (B)

$\frac{t^4}{2}$-$\frac{t^3}{3}$+10t+4

Answer 2

$\alpha= \frac{dω}{dt} =6t^2−2t$

$\int_{0}^{ω}dw=\int_{0}^{t} (6t^2-2t)dt$

So, $\omega = 2t^3 – t^2 \+ 10$

$\int_{4}^{θ} θ dθ = \int_{0}^{t}(2t^3-t^2+10)dt)$

$\theta = \frac{t^4}{2} − \frac{t^3}{3}+10t+4$

$\therefore ,$ The correct option is (B): $\frac{t^4}{2} − \frac{t^3}{3}+10t+4$