Question

A wheel of radius $1\,m$ is spinning with angular velocity which varies with time as $5{t^2}\;rad/s.$ The acceleration of the point of the wheel’s rim at $t = 1\,s,$ has the magnitude equal to

$$A)\;\;5\sqrt {29} \;m/{s^2} \\\ B)\;\;25\;m/{s^2} \\\ C)\;\;10\;m/{s^2} \\\ D)\;\;29\sqrt 5 \;m/{s^2} \\\$$

Answer 1

As the wheel is spinning, so that motion under consideration is rotational motion. To find the acceleration at a point on the rim, relation between linear and angular acceleration is to be used.
$v = rw$
$a = \dfrac{{{v^2}}}{r}$
Where v is the linear velocity
a is the acceleration
r is the radius
w is the angular velocity

Complete step by step answer:
The radius of wheel, $r = 1m$
And the angular velocity of spinning wheel, $w = 5{t^2}$rad/sec
So, angular acceleration, $\alpha = \dfrac{{dw}}{{dt}}$
$\alpha = \dfrac{d}{{dt}}\;(5{t^2})\; \Rightarrow \;\alpha = 10t$
Now, as the wheel is rotating. So, the acceleration is given by
$a = \dfrac{{{v^2}}}{r}$ ….(1)
Where v is the velocity and r is the radius of the wheel.
Now, we know that the relation between linear velocity, v and angular velocity, w is given by
$v = rw$
So,

$$v = (1)\;5{t^2} \\\ v = 5{t^2} \\\$$

at $t = 1\;\sec ,\;\;\;\;v = 5{(1)^2}\;\;\; \Rightarrow \;\;\;v = 5m/\sec$
Putting $r = 1m\;\;\;and\;\;\;v = 5m/\sec$in equation (1), we get
$a = \dfrac{{{{(5)}^2}}}{1} = 25\;m/{\sec ^2}$
So, the acceleration of the point on the wheel’s rim at $t = 1\;\sec$ has magnitude $25\;m/{\sec ^2}$.

So, the correct answer is “Option B”.

Additional Information:
The value of angular acceleration,
$\alpha \;\;at\;t = 1\;\sec \;\;is \\\ \alpha = \;10\;t = 10\;(1) = 10\;rad/{\sec ^2} \\\$
The equation of motion for rotational motion are given by
${w^2} = {w_o}^2 = 2\alpha \theta \\\ w = {w_o} + \alpha t \\\ \theta = {w_o}t + \dfrac{1}{2}\alpha {t^2} \\\$
Where
$\theta$ is angular displacement
$\alpha$ is angular acceleration
w is final angular velocity
${w_o}$ is the time taken.

Note:
Remember that at every point on the wheels rim, the velocity vector acts tangentially. The centripetal acceleration is responsible for circular motion which is $\dfrac{{{v^2}}}{r}$.