Question

a rotating wheel of radius 0.5 m has an angular velocity of 5 rad/sec at some instant and 10 rad/ s after 5 s what is angular acceleration of a point on it's rim

Answer 1

Angular acceleration will be $1rad/sec^2$

Explanation:

We have given initial angular velocity $\omega _i=5rad/sec$

Final angular velocity $\omega _f=10rad/sec$

Time taken for change in velocity t = 5 sec

From first equation of motion we know that

$\omega _f=\omega _i+\alpha t$

$10=5+\alpha\times 5$

$\alpha =1rad/sec^2$

So angular acceleration will be $1rad/sec^2$

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Answer 2

The angular acceleration of a point on its rim is $1\ rad/s^2$.

Explanation:

It is given that,

Radius of the wheel, r = 0.5 m

Initial angular speed of the wheel, $\omega_i=5\ rad/s$

Final angular speed of the wheel, $\omega_f=10\ rad/s$

Time, t = 5 s

To find,

The angular acceleration of a point on it's rim.

Solution,

The angular acceleration can be calculated using first equation of rotational kinematics as :

$\alpha= \dfrac{\omega_f-\omega_i}{t}$

$\alpha= \dfrac{10-5}{5}$

$\alpha =1\ rad/s^2$

So, the angular acceleration of a point on its rim is $1\ rad/s^2$. Hence, this is the required solution.

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