Question

Use the following information for Questions #9 & #10: On a game show, a contestant gives a wheel, appropriately named the Wheel-of-Fortune\, a hard quick shove tangentially to the outside of the wheel with a force of 350N giving it an initial angular speed of 3.3 rad/s. The Wheel-of-Fortune completes 1.75 full revolutions before coming to a stop.

9. What is the magnitude of the angular acceleration of this Wheel of Fortune while it was slowing down uniformly?

a. 0.125 rad/s

b. 0.25 rad/s

c. 0.50 rad/s

d. 0.75 rad/s

10. This Wheel of Fortune is basically a solid disc that has a radius of 1.3 m. What is its mass in kg?

a. 269 kg

b. 539 kg

c. 808 kg

d. 1077 kg

Answer 1

Given
9.
Fortune wheel of radius 1.3 m, applied force tangentially F = 350 N
initial angular speed w1 = 3.3 rad/s
completes 1.75 full revolutions before coming to a stop.

and slowing down uniformly

we can apply the equations of motions

from

w2^2-w1^2 = 2*alpha*theta

final angular velocity is zero that is w2=0 when it comes to stop

alpha = -w1^2 /(2*theta)

here theta is the angular displacement

we know that one revolution the distance covered is equal to 2pi radians
so for 1.7 complete revolutions it covers a distance of 1.7*2pi radians

alpha = -3.3^2/(2*1.7*2pi) rad/s2

alpha = -0.509764 rad/s2

alpha = -0.50 rad/s2 <<<<----------ANSWERR

we know that the -ve sigh indicates the wheel is moving with deceleration slowing down uniformly .

and the unit of angular acceleration is rad/s2 so the answer is option C.0.5 rad/s^2

10.

Given
radius r = 1.3 m
from above the acceleration is alpha = 0.5 rad/s2
Tangential force applied is F = 350 N

due to which torque acts on it that is

   by definition of Torque T = r X F -----(1)
in rotationa motion the trque can be written as T = I* alpha
I is moment of inertia here I = m*r^2/2 ( solid disc)

       T = m*r^2/2* alpha ---------(2)
equating both (1) and (2)

       r*F sin theata = m*r^2/2* alpha

here force applied tangentially so the theta = 90 degrees , sin90 =1

       m = 2*F/(r*alpha ) kg

substituting the values

       m = 2*350 /(1.3*0.5) kg
  
       m = 1076.92 kg

       m = 1077 kg <<<-----------Answer

answer is option (D ) 1077 kg