A merry-go-round starts from rest and accelerates uniformly over11.0 s to a fina

Question

A merry-go-round starts from rest and accelerates uniformly over11.0 s to a final angular velocity of8.40 rev/min. (a) Find the maximum linear speed of a personsitting on the merry-go-round 5.75 m fromthe center.
1 m/s

(b) Find the person's maximum radial acceleration.
2 m/s2

(c) Find the angular acceleration of the merry-go-round.
3rad/s2

(d) Find the person's tangential acceleration.
4m/s2 (a) Find the maximum linear speed of a personsitting on the merry-go-round 5.75 m fromthe center.
1 m/s

(b) Find the person's maximum radial acceleration.
2 m/s2

(c) Find the angular acceleration of the merry-go-round.
3rad/s2

(d) Find the person's tangential acceleration.
4m/s2

Explanation / Answer

(a)The maximum linear speed of a person sitting on themerry-go-round 5.75 m from the centeris v = r * w where r = 5.75 m and w = 8.40 rev/min = 8.40 * (2/60)rad/s = 0.879 rad/s or v = 5.75 * 0.879 = 5.05 m/s (b)Let the person's maximum radial acceleration be a. We know that v = u + at where u = 0 m/s and t = 11.0 s or a = (v - u/t) = (5.05 - 0/11.0) = 0.46m/s2 (c)Let the angular acceleration of the merry-go-round be. We know that w = wo + t where wo = 0 rad/s or = (w - wo/t) = (0.879 - 0/11.0) = 0.079rad/s2 (d)Let the person's tangential acceleration beat. We know that at = r * = 5.75 * 0.079 = 0.45m/s2 (d)Let the person's tangential acceleration beat. We know that at = r * = 5.75 * 0.079 = 0.45m/s2

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