Question

initially at rest a solid body is given an angular acceleration of 15rad/r^2.Point A is 10cm from the rotating axis. After 0.4 second the acceleration of point A (in m/s^2) is?

Answer 1

Hello @dinisha !

Can i know the answer ?

Answer 2

do you mean the options?

Answer 3

yes, or the correct answer if u know

Answer 4

no i don't know!! @Abhisar sorry :(

Answer 5

options ?

Answer 6

don't know even the options ?

Answer 7

A. 1.5
B. 2.1
C. 3.6
D. 3.9
E. 5.1

Answer 8

angular accelaration on every point of a rotating body is constant !
U know this ?

Answer 9

only linear accelaration varies

Answer 10

Linear accelaration = angular accelaration * radial axis

Answer 11

Now can u calculate the answer @dinisha ?

Answer 12

The time given is just to create an illusion !

Answer 13

So there are two components to the acceleration
in rotational motion. There's the tangential acceleration and the centripetal acceleration. We need to figure out both independently. They are at perpendicular to each other, so we can use the pythagorean theorem to get the total acceleration.

For the tangential acceleration, we need to recognize that for rigid bodies, a = r * alpha, where alpha is the angular acceleration. Both r & alpha are given in the problem, so a = r * alpha = 1.5 m/s^2.

For the centripetal acceleration, we have the known result (hopefully known from Newton) a_c = v^2/r.

So again for rigid bodies, v = r * w, where w is the angular velocity. (It's actually an omega not a w).

So simplifying, this means a_c = w^2r.
For constant angular acceleration, w = alpha * t + w_0. But the object starts from rest for w_0 = 0.

Plugging in the appropriate numbers, this simplifies to 3.6 m/s^2.
And then using pythagorean theorem, a = 3.9 m/s^2

Make sense? @dinisha

Answer 14

actually go with her !
\(\color{green}{\huge\ddot\smile}\)

Answer 15

thank you soo much guys.. you all really helped me a lot :) @Abhisar and @Miracrown

Answer 16

\(\Huge\text{Anytime !}\)
\(\huge\ddot\smile\)