Question

The angular acceleration $\alpha$ of a spinning top as a function of $t$ is : $\alpha=3 t^{2}+5 t$. At $t=0$, the angular velocity $\omega_{0}=10 rad / s$. and angular position $\theta_{0}=8 rad$. The angular position as a function of time $t$ is given by which of the following expression?

• A. $\frac{1}{4} t^{4}+\frac{5}{6} t^{3}+10 t+8$
• B. $\frac{5}{6} t^{4}+\frac{1}{4} t^{3}+\frac{2}{5} t+8$
• C. $2 t^{4}+3 t^{3}+5 t+8$
• D. $\frac{1}{4} t^{4}+\frac{3}{5} t^{3}+6 t^{2}+8$

Answer 1

Answer: (A)

Given : $\alpha=3 t^{2}+5 t$
$\alpha=\frac{d^{2} \theta}{d t^{2}}=3 t^{2}+5 t \,\,\,\,\, ...(i)$
Integrate both sides w.r.t. $t$ in equation $(i)$ , we get
$\frac{d \theta}{d t}=t^{3}+\frac{5}{2} t^{2}+C_{1}$
where $C_{l}$ is a constant of integration.
or $ \omega=\frac{d \theta}{d t}=t^{3}+\frac{5}{2} t^{2}+C_{1}$
Using initial conditions, at $t=0, \omega_{0}=10 \,rad / s$
$\therefore C_{I}=10\, rad / s$
or $ \frac{d \theta}{d t}=t^{3}+\frac{5}{2} t^{2}+10 \,\,\,\,\,\, ....(ii)$
Integrate both sides w.r.t. $t$ in equation $(ii)$ , we get
$\theta=\frac{1}{4} t^{4}+\frac{5}{6} t^{3}+10 t+C_{2}$
where $C_{2}$ is a constant of integration.
Using initial conditions, at $t=0, \theta_{0}=8 \,rad$
$\therefore C_{2}=8\, rad $
or $ \theta=\frac{1}{4} t^{4}+\frac{5}{6} t^{3}+10 t+8$