(IV) A merry-go-round consists of a solid disc (platform), a thin railing, and 1

Question

(IV) A merry-go-round consists of a solid disc (platform), a thin railing, and 14 horses, ar- ranged in a two rows. The platform has mass My 400 kg and radius RP = 3m. The rail- ing with total mass M100 kg runs along the edge of the platform, so R 3m Each of the horses has mass MA 25 kg. Six of the horses are at inner radius Ri1n, the other eight are at outer radius R-2m o. and T he merry-go-round rotates rigidly about the axis through its centre. At t 0 the merry 8o-round is initially at rest and it begins to accelerate at 1/3 rad/e2. This acceleration lasts until t = 68 at which it drops to zero, and the merry-go-round continues to rotate at a constant angular velocity (a) (6) Determine the angular velocity of the merry-go-round in the time interval 0sts6s (i) Determine the angular velocity of the merry-go-round for times after t 6s (ii) Determine the angular displacement (from its rest position) during 0ss6a (iv) Determine the angular displacement for times after 6s. (c) ) Modelling the platform as a uniform circular disc, compute its moment of inertia. (ii) Modelling the railing as a thin ring, compute its moment of inertia. (ii) Modelling each inner horse as a point mass, compute the moment of inertia of one inner horse. (iv) Modelling each outer horse as a point mass, compute the moment of inertia of one outer horse. (v) Compute the total moment of inertia of the merry-go-round. (d) Determine the value of the net torque labout the axis of symmetry) acting on the merry- go-round. (e) Suppose that a wheel under the platform, at distance r -2.5a from the centre, drives the merry-go-round with a force of 600N, causing it to accelerate (during the first six seconds). Estimate the torque provided by the frictional forcels)that act on the bearings that rub on the axle about which the system rotates. Page 4 of 4

Explanation / Answer

IV) merry go round

solid disc (platform) , mass Mp = 400 kg

radius, Rp = 3m

Mr = 100 kg ( on the radial distacne of the center)

Rr = 3 m

Mh = 25 kg

total numebr of horses, n = 14

Ri = 1 m ( 6 horses)

Ro = 2 m ( 8 horses)

a. at t = 0, initial angular speed = wo = 0

angular acceleration alpha = (1/3) rad/s/s

t = 6 s

after this the merry go round continures to rotate at a constant angular velocity

so total moment of inertia of the whoel system = I

I = 0.5Mp*Rp^2 + Mr*Rr^2 + 8*Mh*Ro^2 + 6*Mh*Ri^2

I = 3650 kg m^2

hence for , 0 < t < 6

w = wo + alpha*t

w = 0 + (1/3)t = t/3

ii. after t = 6 s

w(6) = 2 rad/s

iii. angular position = theta

for 0 < t < 6

theta = 0.5*alpha*t^2 = t^2/6

iv) after t = 6

theta = 36/6 + (t - 6)w(6) = 6 + (t - 6)2 = 2t - 6 rad

c. i.moment of inertia of uniform disc = 0.5MpRp^2 = 1800 kg m^2

ii. moment of inertia of ring = Mr*Rr^2 = 900 kg m^2

iii. moment of inertia of one inner horde = Mh*ri^2 25kg m^2

iv. momwent of inertia of one outer hors = Mh*ro^2 = 100 kg m^2

v. total moment of inertia = 3650 kgm^2

d. net torque = alpha*I = 1216.66666666 Nm

e. d = 2.5 m

F = 600 N

frictional torque = f'

then

Fd - f' = 1216.666666666

f' = 283.33333333333333 Nm

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