Question

The angular acceleration $\alpha$ of a spinning top as a function of $t$ is : $\alpha =3t^2+5t$. At $t=0$, the angular velocity ${\omega }_0=10rad/s$. and angular position ${\theta }_0=8rad$. The angular position as a function of time $t$ is given by which of the following expression?

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

Answer 1

Answer: (A)

Given : $\alpha =3t^2+5t$
$\alpha =\frac{d^2\theta }{d{t}^2}=3t^2+5t...(i)$
Integrate both sides w.r.t. $t$ in equation $(i)$ , we get
$\frac{d\theta }{dt}=t^3+\frac{5}{2}{t}^2+C_1$
where $C_l$ is a constant of integration.
or $\omega =\frac{d\theta }{dt}=t^3+\frac{5}{2}{t}^2+C_1$
Using initial conditions, at $t=0,{\omega }_0=10rad/s$
$\therefore C_I=10rad/s$
or $\frac{d\theta }{dt}=t^3+\frac{5}{2}{t}^2+10....(ii)$
Integrate both sides w.r.t. $t$ in equation $(ii)$ , we get
$\theta =\frac{1}{4}{t}^4+\frac{5}{6}{t}^3+10t+C_2$
where $C_2$ is a constant of integration.
Using initial conditions, at $t=0,{\theta }_0=8rad$
$\therefore C_2=8rad$
or $\theta =\frac{1}{4}{t}^4+\frac{5}{6}{t}^3+10t+8$