Question

A bar of length L and mass m has a frictionless pivot through its mid point. There is an additional point
mass 2m on the right end of the bar and an additional point mass m on the left end of the bar. The bar
is held in horizontal position by a vertical cord attached at L/4 from the left end as shown in the Figure.
The additional masses m & 2m are fixed to the rod. The rod is initially horizontal. Find the force (in
Newton) that the pivot exerts on the bar (if m = 1 kg and L = 0.5m).​

Answer 1

Answer:

PHYSICS

A bar of length L and mass M has a frictionless pivot through its mid point. An additional small block of mass 2M is attached on the right end of the bar and small block of mass M is attached on the left end of the bar. The bar is held in horizontal position by a vertical cord attached to a point at distance L/4 from the left end as shown in the figure. When the bar has been rotated through 90

o

and is vertical, what is the speed of the mass 2M?

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ANSWER

ΔU In Process =−2m

2

L

g+m

2

L

g=−m

2

L

g

ΔU+ΔK=0⇒ΔK=−ΔU

⇒ΔK=m

2

L

g

2

1

Iω

2

=m

2

L

g⇒

2

1

.

6

5

mL

2

ω

2

=m

2

L

g

⇒ω=

5L

6g

Velocity of 2m=

2

L

ω=

k

3gL

solution

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