A bob of mass m , suspended by a string of length l1 , is given a minimum velocity required to complete a full circle in the vetical plane .At the highest point ,it collides elastically with another bob of mass m suspended by a string l2 , which is initially at rest.Both the strings are massless and inextensible.If the second bob,after collision acquires the minimum speed required to complete full vertical circle. Find l1/l2 .
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Answered by
45
See the diagram.
For a bob at the top position in the circle, the forces are centripetal force, weight and tension.
T = m v² / r - m g --- (1)
String remains tight if T > 0. Hence, v² >= r g --- (2)
Let velocity given to the bob at its lowest position in the circle = u.
1/2 m u² = 1/2 m v² + m g (2 r)
=> u² = v² + 4 g r
=> u >= √(5 rg) --- (3) using (2)
Now for the Top bob tied to string of radius L2:
Velocity given at it lowest position = v2 >= √(5 L2 g) -- (4)
The collision of two bobs of the same mass m is elastic, and the bob tied to string of length L2, was at rest before collision, we apply conservation of linear momentum and energy. We get
v1 = v2 >= √(5 L2 g) --- (5)
Velocity of bob tied to string of length L1 is same as that of the other bob after collision. Then after collision, the first bob's velocity is 0.
Let the velocity of first bob at its lowest position = v0. Apply conservation of energy:
1/2 m v0² = 1/2 m v1² + m g (2 L1)
vo² = v1² + 4 g L1
= 5 g L2 + 4 g L1 --- (6)
As the first bob was given just enough velocity v0 at its lower position to make a full circle,
v0² >= 5 L1 g using (3)
= 5 g L2 + 4 g L1 from (6)
Solve these two to get ans.
Hence, L1/ L2 = 5
For a bob at the top position in the circle, the forces are centripetal force, weight and tension.
T = m v² / r - m g --- (1)
String remains tight if T > 0. Hence, v² >= r g --- (2)
Let velocity given to the bob at its lowest position in the circle = u.
1/2 m u² = 1/2 m v² + m g (2 r)
=> u² = v² + 4 g r
=> u >= √(5 rg) --- (3) using (2)
Now for the Top bob tied to string of radius L2:
Velocity given at it lowest position = v2 >= √(5 L2 g) -- (4)
The collision of two bobs of the same mass m is elastic, and the bob tied to string of length L2, was at rest before collision, we apply conservation of linear momentum and energy. We get
v1 = v2 >= √(5 L2 g) --- (5)
Velocity of bob tied to string of length L1 is same as that of the other bob after collision. Then after collision, the first bob's velocity is 0.
Let the velocity of first bob at its lowest position = v0. Apply conservation of energy:
1/2 m v0² = 1/2 m v1² + m g (2 L1)
vo² = v1² + 4 g L1
= 5 g L2 + 4 g L1 --- (6)
As the first bob was given just enough velocity v0 at its lower position to make a full circle,
v0² >= 5 L1 g using (3)
= 5 g L2 + 4 g L1 from (6)
Solve these two to get ans.
Hence, L1/ L2 = 5
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Answered by
8
Mini. Velocity required for first bob to compleate circle= √5gl1
Velocity at upper end = √gl1
Mini.velocity required for 2nd bob to compleate circle=√5gl2
After collide upper end velocity of 1st bob = velocity required for compleate circle of 2nd bob
So,
√gl1 =√5gl2
L1/L2 = 5
Thanks.....
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