11.) A 210-kg merry-go-round in the shape of a uniform, solid, horizontal disk o

Question

11.) A 210-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s? (State the magnitude of the force.)

12.) A car is designed to get its energy from a rotating flywheel with a radius of 1.50 m and a mass of 400 kg. Before a trip, the flywheel is attached to an electric motor, which brings the flywheel's rotational speed up to 5,650 rev/min.

(a) Find the kinetic energy stored in the flywheel. (J)

(b) If the flywheel is to supply energy to the car as a 20.0-hp motor would, find the length of time the car could run before the flywheel would have to be brought back up to speed. (h)

13.) A light rod of length = 1.00 m rotates about an axis perpendicular to its length and passing through its center as in the figure below. Two particles of masses m1 = 4.35 kg andm2 = 3.00 kg are connected to the ends of the rod.

(a) Neglecting the mass of the rod, what is the system's kinetic energy when its angular speed is 2.00 rad/s?  (J)

(b) Repeat the problem, assuming the mass of the rod is taken to be 2.15 kg. (J)

Explanation / Answer

11)
Given that
initial angular speed is wi = 0 rad/s

final angular speed is wf = 0.6*2*3.142 rad/s = 3.7704 rad/s

time taken is t = 2S

angular accelaration alpha = (wf-wi)/t = (3.7704-0)/2 = 1.8852 rad/s^2

allpy torque T = I*alpha

I is the moment of inertia =0.5*m*r^2 = 0.5*210*1.5^2 = 236.25 kg*m^2

then T = r*F = 1.5*F

1.5*F = 236.25*1.8852
F = 297 N
=======================================================

12)

Kinetic energy stored is 0.5*I*w^2 = 0.5*0.5*400*1.5^2*(5650*2*3.142/60)^2 = 7.87*10^7 J

B) power P = Energy/time = E/t

time t = E/P = (7.87*10^7)/(20*746) = 5274.8 S = 1.46 h

=========================================================

13) momnt od inertia of the sysytem isI = (1/12)*M*L^2 + (m1*r^2)+(m2*r^2) = (1/12)*0*L^2 + (4.35*0.5^2)+(3*0.5^2)

I = 1.8375 kg*m^2

kinetic energy of the system is = 0.5*I*w^2 = 0.5*1.8375*2^2 = 3.675 J

B) Noe new I = (1/12)*M*L^2 + (m1*r^2)+(m2*r^2)

I = (1/12)*(2.15)*(1)^2 + (4.35*0.5^2) + (3*0.5^2)
I = 2.01 kg*m^2

Kinetic energy KE = 0.5*I*w^2 = 0.5*2.01*2^2 = 4.02 J

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