1. A person exerts a force of 43.9N on the end of a door 79.6cm wide. What is th

Question

1. A person exerts a force of 43.9N on the end of a door 79.6cm wide. What is the magnitude of the torque if the force is exerted at a 64.4o angle to the face of the door?

2. a.A day-care worker pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a spinning rate of 21.7 rpm in 10.6s. Assume the merry-go-round is a disk of radius 2.36m and has a mass of 710kg, and two children (each with a mass of 22.5kg) sit opposite each other on the edge. Calculate the torque required to produce the acceleration, neglecting frictional torque.

    b.What force is required?

Explanation / Answer

1) F =43.9 N , d = 79.6 cm , theta =64.4 degrees

Torque = rxF =rFsin(theta) = 0.796*43.9*sin(64.4)

Torque = 31.514 N.m

2) a) wo =0 ,w = 21.7 rpm = (21.7*2*3.14)/60 = 2.27 rad/s

t= 10.6s

from rotational kinematic equation

w =wo+ (aplha)t

angular acceleration alpha = 2.27/10.6 = 0.214 rad/s^2

R = 2.36 m , M =710 kg , m = 22.5 kg

Moment of inertia of disk = MR^2/2

Total moment of inertia I = (MR^2/2) +(mR^2)+(mR^2)

I = (710*2.36*2.36/2) +(2*22.5*2.36*2.36)

I = 2227.8 kg.m^2

Torque = I *aplha = 2227.8*0.214

Torque = 476.75 N.m

b) Torque = RxF

476.75 = 2.36*F
F = 202 N

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