You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Angular Impulse

There are two...

Question

There are two point masses each of mass m are at the ends of a rod of mass m. The rod is released. What will be the angular speed when it becomes vertical?

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B

Moment of internal of the rod about refernce point (i)
$m (\frac{L}{4})^{2} + [\frac{m L^{2}}{12} + m (\frac{L}{4})^{2}] + m (\frac{3 L}{4})^{2}$

$= \frac{37 m L^{2}}{48}$
Basing on energy conservation,
$K_{i} + P_{i} = K_{f} + P_{f}$
$0 + 0 = \frac{l ω^{2}}{2} + P_{f}$

Suggest Corrections

Similar questions

Q. Two masses each of

m

are attached at mid point

B

& end point

C

of massless rod

A C

which is jinged at

A

. It is released from horizontal position as shown. Find the force at hinge

A

when rod becomes vertical

A composite rod of mass '2m' $&$ length '2l' consists of two identical rods joined end to end at P. The composite rod is hinged at one of its ends and is kept horizontal. If it is released from rest, ·Find its angular speed when it becomes vertical

Q. A rigid rod of mass

m

& length

ℓ

is pivoted at one of its ends. If it is released from its horizontal position, find the speed of the centre of mass of the rod when it becomes vertical.

Q. A composite rod of mass

2 m

and length

2 l

comprises two identical rods joined end to end at

P

. The composite rod hinged at one of its end is kept horizontal as shown in the figure. If it is released from rest, find its angular speed when it becomes vertical.

Join BYJU'S Learning Program

There are two point masses each of mass m are at the ends of a rod of mass m. The rod is released. What will be the angular speed when it becomes vertical?

The correct option is B Moment of internal of the rod about refernce point (i) m(L4)2+[mL212+m(L4)2]+m(3L4)2 =37mL248 Basing on energy conservation, Ki+Pi=Kf+Pf 0+0=lω22+Pf

The correct option is B

Moment of internal of the rod about refernce point (i)
$m (\frac{L}{4})^{2} + [\frac{m L^{2}}{12} + m (\frac{L}{4})^{2}] + m (\frac{3 L}{4})^{2}$

$= \frac{37 m L^{2}}{48}$
Basing on energy conservation,
$K_{i} + P_{i} = K_{f} + P_{f}$
$0 + 0 = \frac{l ω^{2}}{2} + P_{f}$