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प्रश्न
Using the energy conservation, derive the expressions for the minimum speeds at different locations along a vertical circular motion controlled by gravity. Is zero speed possible at the uppermost point? Under what condition/s?
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उत्तर
Consider a small body (or particle) of mass m tied to a string and revolved in a vertical circle of radius r at a place where the acceleration due to gravity is g. At every instant of its motion, the body is acted upon by two forces, namely,
its weight `vec(mg)` and
the tension `vecT` in the string.
(i) At the top (point A): Let v1 be the speed of the particle and T1 the tension in the string. Here, both `vec(T_1)` and weight `vec(mg)` are vertically downward. Hence, the net force on the particle towards the centre O is T1 + mg, which is the necessary centripetal force.
`therefore T_1 + mg = (mv_1^2)/r` ...(1)
To find the minimum value of v1 that the particle must have at the top, we consider the limiting case when the tension T1 just becomes zero.
`therefore (mv_1^2)/r = mg`
that is, the particle's weight alone is the necessary centripetal force at point A.
`therefore v_1^2 = gr`
`therefore v_1 = sqrt(gr)` ...(2)
 |
| Vertical circular motion (schematic) |
(ii) At the bottom (point B): Let v2 be the speed at the bottom. Taking the reference level for zero potential energy to be the bottom of the circle, the particle has only kinetic energy `1/2 mv_2^2` at the lowest point.
Total energy at the bottom = KE + PE
= `1/2 mv_2^2 + 0 = 1/2 mv_2^2` ...(3)
As the particle goes from the bottom to the top of the circle, it rises through a height h = 2r. Therefore, its potential energy at the top is
mgh = mg (2r)
and, from Eq. (2), its minimum kinetic energy there is
`1/2 mv_1^2 = 1/2 mgr` ...(4)
Minimum total energy at the top = KE + PE
= `1/2 mgr + 2 mgr`
= `5/2 mgr` ...(5)
Assuming that the total energy of the particle is conserved, total energy at the bottom = total energy at the top. Then, from Eqs. (4) and (5),
`1/2 mv_2^2 = 5/2 mgr`
The minimum speed the particle must have at the lowest position is
`v_2 = sqrt(5 gr)` ...(6)
(iii) At the midway (point C): Let v3 be the speed at point C, so that its kinetic energy is `1/2 mv_3^2`.
At C, the particle is at a height r from the bottom of the circle. Therefore, its potential energy at C is mgr.
Total energy at C = `1/2 mv_3^2 + mgr` ...(7)
From the law of conservation of energy,
total energy at C = total energy at B
`therefore 1/2 mv_3^2 + mgr = 5/2 mgr`
`therefore v_3^2 = 5gr - 2gr`
`therefore v_3^2 = 3gr`
∴ The minimum speed the particle must have midway up is `v_3 = sqrt(3 gr)` ...(8)
In non-uniform vertical circular motion, e.g., those of a small body attached to a string or the loop-the-loop manoeuvres of an aircraft, motorcycle, or skateboard, the body must have some minimum speed to reach the top and complete the circle. In this case, the motion is controlled only by gravity, and zero speed at the top is not possible.
However, in a controlled vertical circular motion, e.g., those of a small body attached to a rod or the giant wheel (Ferris wheel) ride, the body or the passenger seat can have zero speed at the top, i.e., the motion can be brought to a stop.
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संबंधित प्रश्न
Answer in Brief:
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