#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# a uniform rod of length l,hinged at lower end is free to rotate in the vertical plane.if the rod is held vertically in the beginning and then released,the angular acceleration of the rod when it makes and angle 45 deg. with horizontal is?

147 Points
11 years ago

Dear tushar

When it makes 45 degree with the horizontal then at this position torque acting on rod is =mgl/2 cos45

=mgl/2√2

We know that

Torque=moment of inertia * angular acceleration

mgl/2√2  =ml2/3  * angular acceleration         where m is the mass of rod

angular acceleration =3g/2√2 l

Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.

All the best.

Regards,

Rishi Sharma
11 months ago
Dear Student,

When it makes 45 degree with the horizontal then at this position torque acting on rod is
=mgl/2 cos45
=mgl/2√2
We know that
Torque=moment of inertia * angular acceleration
= mgl/2√2
= ml2/3 * angular acceleration
where m is the mass of rod
angular acceleration =3g/2√2 l

Thanks and Regards