A tiny block is free to slide over the frictionless surface of a hemispherical b

Question

A tiny block is free to slide over the frictionless surface of a hemispherical bowl, of radius r. The block is displaced a small distance from the bottom of the bowl and released.

x(t) = xmax cos(?t + ?)

? = sqrt (g/r)

If the radius of the bowl is r = 30 cm, find the period of the block’s oscillation. (Solved this, T = 1.1s)

(Need help with this one) After the block has begun oscillating, a timer is started at t = 0 just when the block reaches a point x = ?2 cm (to the left of the bottom of the bowl), moving rightward at a speed of 9 cm/s. What is the amplitude of the motion xmax and what is the phase angle ? for this choice of starting time?

Explanation / Answer

T = 2pi / w

w = sqrt(g/r) = sqrt(9.8 / 0.30) = 5.71 rad/s

T = 2pi / 5.71 = 1.10 s .....Ans (time period)

at t = 0

x= -2 cm = - 0.02 m

spring PE = kx^2 /2 = k (0.02)^2 /2 = 0.0002k

KE = mv^2 /2 = m (0.09)^2 /2 = 0.00405m

total energy = PE + KE = 0.0002k + 0.00405m

and mw^2 = k

m = k / (5.71^2)   = 0.0307k

TE = 0.0002k + (0.00405 x 0.0307k) = 0.000324k

when object is at extreme position,

say x = A

then KE = 0

so total energy = 0 + kA^2 /2 = 0.000324k

A = 0.0255 m = 2.55 cm

Xmax = A = 2.55 cm .............Ans

x(t) = 2.55 cos(5.71t + phi )

at t= 0 , x = -2 cm

-2 = 2.55 cos(0 + phi)

cos(phi) = - 0.784

phi = 2.47 rad ...........Ans

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