Question

A point on a wheel describes an angular distance theta=2t^2. find the time after which the radial and tangential acceleration of the point will be numerically equal.​

Answer 1

Given:-

Angular distance, $\theta=2t^{2}$

To Find:-

time at which radial and tangential acceleration acceleration becomes equal.

Explanation:-

$\text{Angular velocity, }\omega=\dfrac{d\theta}{dt}=4t\\\\\text{Angular acceleration, }\alpha=\dfrac{d\omega}{dt}=4\\\\\text{Now, }\\\\\text{radial acceleration, }a_{r}=r\times\omega^{2}=8rt^{2}\\\\\text{tangential acceleration, }a_{t}=r\times\alpha=4r\\\\\text{ATQ,}\\a_{t}=a_{r}\\\\\Rightarrow 8rt^{2}=4r\Rightarrow t^{2}=\dfrac{1}{2}\Rightarrow t=\dfrac{1}{\sqrt{2}}=0.707s$