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KENNIESAWSTATE UNIVERSTTY PITYS 1T FALL SE MI STER2017 Frp-e ass Exam # 4: 1/16/

ID: 1786238 • Letter: K

Question

KENNIESAWSTATE UNIVERSTTY PITYS 1T FALL SE MI STER2017 Frp-e ass Exam # 4: 1/16/2017 at 5:00 PM Student Name: Student 1D MULTIPLE CHOICE PROBILEMS 5. (15 pts) A wheel of radius 0.60 rotates with a constant angular speed about an axis perpendicular to its center. A point on the wheel that is 0.40 m from the center has a tangential speed of 3.2 mis. Calculate (a) the angular speed of the wheel, (b) the tangential speed of a point 0.5 m from the center the wheel, and (c) the magnitude of the total acceleration of the point that is 0.6 m from the center t. (1o pts) A wheel, initially at rest, is given a constant angular acceleration so that it makes 20.0 revolutions in the first 8.0o s. What is its angular acceleration? a. 0.313 rad/s b. 0.625 rad/s of c. 1.97 rad/s d. 2.50 rad/s e. 3,93 rad/s2 2. (10 pts) A fan rotating with an initial angular velocity of 1000 rev/min is switched off. In 4 seconds, the angular velocity decreases to 800 rev/min. Assuming the angular acceleration is constant, how many revolutions does the blade undergo during this time? a. 10 b. 20 c. 60 d. 125 e. 1200 3. (10 pts) A wheel rotating about a fixed axis with a constant angular acceleration of 2.2 rad/s turns hrough 2.4 revolutions during a 2.0-s time interval What is the angular velocity at the end of this time interval? a. 9.7 rad/s b. 9.5 rad/s c. 9.3 rad/s d. 9.1 rad/s e. 8.8 rad/s 4. (10 pts) A 66-N m torque acts on a wheel with a moment of inertia 275 kg m2. If the wheel starts from rest, how long will it take the wheel to make one revolution? a. 0.37 s c. 2.4 s d. 5.8 s e. 7.2s

Explanation / Answer


1)

initial angular speed wo = 0

angular displacement theta = 20 revolutions = 20*2pi radians


time interval t = 8 s


from equation of rotatory motion

theta = wo*t + (1/2)*alpha*t^2

20*2*pi = 0 + (1/2)*alpha*8^2

alpha = 3.93 rad/s^2 <<<-----------ANSWER

OPTION ( e )

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2)

initial angular speed wo = 1000 rev/min = 1000/60 rev/s

angular displacement theta = ?


time interval t = 4s

final speed w = 800 rev/min = 800/60 rev/s


from equation of rotatory motion

theta = (wo + w)*t/2

theta = (1000/60 + 800/60)*4/2


theta = 60 revolutions

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3)


angular displacemnet , theta = wi*t +( 1/2)*alpha*t^2


2.4*2pi = wi*2 + (1/2)*2.2*2^2


wi = 5.34 rad/s


wf = wi + alpha*t


wf = 5.34 + (2.2*2)

wf = 9.7 rad/s <<<<------ANSWER


OPTION (a)

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(4)


torque = I*alpha


66 = 275*alpha


alpha = 0.24 rad/s^2


theta = wo*t + (1/2)*alpha*t^2


2*pi = 0 + (1/2)*0.24*t^2


t = 7.2 s <<<<-----------------ANSWER