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

Byju's Answer

Standard XII

Physics

The 'e' Equation

A uniform bar...

Question

A uniform bar of mass $M$ and length $L$ collides with a horizontal surface. Before collision, velocity of centre of mass was $v_{0}$ and no angular velocity. Just after collision, velocity of centre of mass of bar becomes $v$ in upward direction as shown. Angular velocity $ω$ of the bar just after impact is:

\frac{3 (v_{0} + v) c o s θ}{2 L}

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

\frac{6 (v_{0} - v) c o s θ}{L}

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

\frac{(v_{0} + v) c o s θ}{L}

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

\frac{(v_{0} - v) c o s θ}{6 L}

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

Open in App

Solution

The correct option is A $\frac{3 (v_{0} + v) c o s θ}{2 L}$
Angular momentum will be conserved about point A just before and just after collision.
$m v_{0.} \frac{L}{2} c o s θ = \frac{m L^{2}}{3} . ω - m v . \frac{L}{2} c o s θ$
$ω = \frac{3 (v_{0} + v) c o s θ}{2 L} .$

Suggest Corrections

Similar questions

Q. A uniform bar of length

6 a

and mass

8 m

lies on a smooth horizontal table. Two point masses

m

and

2 m

, moving in the same horizontal plane with speeds

2 v

and

v

respectively strike the bar (as shown in figure) and stick to the bar after collision. By denoting angular velocity (about centre of mass), total energy and velocity of centre of mass after collision as

ω, E

and

v_{o}

respectively, we can say that

Q. A bar of mass

M

and length

L

is in pure translatory motion and its centre of mass has velocity

V

. It collides and sticks to a second identical bar which is initially at rest. (Assume that it becomes one composite bar of length

2 L

). The angular velocity of the composite bar after collision will be :

Q. A uniform rod of mass

m

and length

l

is free to rotate about a fixed horizontal axis passing through hinge. A small ball of mass

m

moving horizontally with velocity

v_{o}

strikes the rod at depth

y

from hinge as shown in figure. If angular velocity of rod and velocity of ball just after the collision, be

ω

and

v

respectively, then which of the following statements are wrong?

Q. A uniform bar of length

6 a

and mass

8 m

lies on a smooth horizontal table. Two point masses

m

and

2 m

moving in opposite direction in the same horizontal plane with speed

2 v

and

v

, respectively. strike the bar and stick to the bar after the collision. Calculate :
The angular velocity of the centre of mass after collision.

A uniform bar of length 6a and mass 8 m lies on a smooth horizontal table. Two point masses m and 2m moving in the same horizontal plane with speed 2v and v respectively, strike the bar and stick to the bar after collision. Denoting angular velocity (about the centre of mass), total energy and centre of mass velocity by $ω$ , $E$ and $v_{c}$ respectively, we have after collision

Join BYJU'S Learning Program

A uniform bar of mass M and length L collides with a horizontal surface. Before collision, velocity of centre of mass was v0 and no angular velocity. Just after collision, velocity of centre of mass of bar becomes v in upward direction as shown. Angular velocity ω of the bar just after impact is:

The correct option is A 3(v0+v)cosθ2LAngular momentum will be conserved about point A just before and just after collision.mv0.L2cosθ=mL23.ω−mv.L2cosθω=3(v0+v)cosθ2L.

The correct option is A $\frac{3 (v_{0} + v) c o s θ}{2 L}$
Angular momentum will be conserved about point A just before and just after collision.
$m v_{0.} \frac{L}{2} c o s θ = \frac{m L^{2}}{3} . ω - m v . \frac{L}{2} c o s θ$
$ω = \frac{3 (v_{0} + v) c o s θ}{2 L} .$