A small sphere of m = 0.12 kg is suspended by a light string from a vertical pos

Question

A small sphere of m = 0.12 kg is suspended by a light string from a vertical post mounted at the edge of a merry-go-round of radius R = 0.41 m. The equilibrium position of the sphere is shown in the figure, with `a' = 0.45 m and `b' = 0.82 m.

The axis of rotation is indicated by the dashed line.

Calculate the angular velocity of the merry-go-round.

A small sphere of m = 0.12 kg is suspended by a light string from a vertical post mounted at the edge of a merry-go-round of radius R = 0.41 m. The equilibrium position of the sphere is shown in the figure, with 'a' = 0.45 m and 'b' = 0.82 m. The axis of rotation is indicated by the dashed line.

Explanation / Answer

m = 0.120 kg, R = 0.41 m. `a' = 0.45 m and `b' = 0.82 m
Calculate the velocity v of the merry-go-round.
the radius of the sphere = r = R + a
angle of the string with vertical ?, where tan? = a/b

T = tension of the string

vertical net force = Tcos? - mg = 0, so T = mg/cos?

horizontal net force = Tsin? = mv2/r

v2 = Trsin?/m = grtan? = g(R + a)*a/b

?v = ?[g(R + a)*a/b] = sq(9.81*(0.41+0.45)*0.45/0.82)

2.222273061749075 m/s

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