A block of mass m is moving on a smooth horizontal surface with a speed v. In the path of the block there is a fixed disc, free to rotate about its center as shown in the figure. Speed of the block after passing over the disc becomes. If the mass of the disc is also m and radius is R then find angular velocity of the disc.
Answers & Comments
The problem described requires the application of the laws of conservation of linear momentum and conservation of angular momentum.
Conservation of linear momentum states that the total linear momentum of a system is conserved if no external forces act on it.
Conservation of angular momentum states that the total angular momentum of a system is conserved if no external torques act on it.
The change in linear momentum of the block-disc system can be found using the equation Δp = mv - mv', where m is the mass of the block, v is its initial speed, and v' is its final speed.
The change in angular momentum of the disc can be found using the equation ΔL = IΔω, where I is the moment of inertia of the disc and Δω is the change in angular velocity.
By equating the change in linear momentum to the external impulse applied to the system and equating the change in angular momentum to the external torque applied to the system, the final velocity of the block and the angular velocity of the disc can be determined.
Unfortunately, without more specific information regarding the masses, speeds, and sizes involved, it is not possible to provide a numerical solution to this problem.
Answer:
Correct option is A)
The angular momentum of mass about center will remain constant as the torque of tension is zero.
mv
∘
r
∘
=mv
′
2
r
∘
v
′
=2v
∘
Final value of K.E.=
2
1
m(2v
∘
)
2
=2mv
∘
2
Video ExplanationCorrect option is A)
The angular momentum of mass about center will remain constant as the torque of tension is zero.
mv
∘
r
∘
=mv
′
2
r
∘
v
′
=2v
∘
Final value of K.E.=
2
1
m(2v
∘
)
2
=2mv
∘
2
Video ExplanationCorrect option is A)
The angular momentum of mass about center will remain constant as the torque of tension is zero.
mv
∘
r
∘
=mv
′
2
r
∘
v
′
=2v
∘
Final value of K.E.=
2
1
m(2v
∘
)
2
=2mv
∘
2
Video ExplanationCorrect option is A)
The angular momentum of mass about center will remain constant as the torque of tension is zero.
mv
∘
r
∘
=mv
′
2
r
∘
v
′
=2v
∘
Final value of K.E.=
2
1
m(2v
∘
)
2
=2mv
∘
2
Video Explanation