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प्रश्न
Answer the following question.
Show that the centripetal force on a particle undergoing uniform circular motion is -mrω2.
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उत्तर
- Suppose a particle is performing U.C.M in anticlockwise direction. The co-ordinate axes are chosen as shown in the figure.
Let, A = initial position of the particle which lies on positive X-axis
P = instantaneous position after time t
θ = angle made by radius vector
ω = constant angular speed
`vec"r"` = instantaneous position vector at time t - From the figure,
`vec"r" = hat"i""x" + hat"j""y"`
where, `hat"i" and hat"j"` are unit vectors along X-axis and Y-axis respectively.
- Also, x = r cos θ and y = r sin θ
∴ `vec"r" = ["r"hat"i"cos theta + "r"hat"j"sin theta]`
But θ = ωt
∴ `vec"r" = ["r"hat"i" cos omega"t" + "r"hat"j" sin omega"t"]` ....(1) - Velocity of the particle is given as rate of change of position vector.
∴ `vec"v" = (vec"dr")/"dt" = "d"/"dt" ["r"hat"i" cos omega"t" + "r" hat"j" sin omega"t"]`
`= "r"["d"/"dt" cos omega"t"]hat"i" + "r"["d"/"dt" sin omega"t"]hat"j"`
∴ `vec"v" = - "r"omega hat"i" sin omega"t" + "r"omega hat"j" cos omega"t"`
∴ `vec"v" = "r"omega (-hat"i" sin omega"t" + hat"j" cos omega"t")` - Further, instantaneous linear acceleration of the particle at instant t is given by,
`vec"a" = (vec"dv")/"dt" = "d"/"dt"["r"omega(- hat"i" sin omega "t" + hat"j" cos omega"t")]`
`= "r"omega ["d"/"dt" (- hat"i" sin omega"t" + hat"j" cos omega "t")]`
`= "r"omega ["d"/"dt" (- sin omega"t")hat"i" + "d"/"dt"(cos omega "t")hat"j"]`
`= "r"omega(-omega hat"i" cos omega"t" - omega hat"j" sin omega "t")`
`= - "r"omega^2(hat"i" cos omega "t" + hat"j" sin omega"t")`
∴ `vec"a" = - omega^2("r"hat"i" cos omega"t" + "r" hat"j" sin omega "t")` ....(2) - From equation (1) and (2),
`vec"a" = - omega^2vec"r"` ....(3)
The negative sign shows that the direction of acceleration is opposite to the direction of the position vector. Equation (3) is the centripetal acceleration. - The magnitude of centripetal acceleration is given by,
a = `omega^2"r"` - The force providing this acceleration should also be in the same direction, hence centripetal.
∴ `vec"F" = "m"vec"a" = - "m"omega^2vec"r"`
This is the expression for the centripetal force on a particle undergoing uniform circular motion. - Magnitude of F = `"m"omega^2"r" = "mv"^2/"r"` = mωv
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