q9.10 A flywheel with a radius of 0.600 m starts from rest and accelerates with

ID: 1396175 • Letter: Q

Question

q9.10

A flywheel with a radius of 0.600 m starts from rest and accelerates with a constant angular acceleration of 0.200 rad/s2 .

Part A

Compute the magnitude of the tangential acceleration of a point on its rim at the start.

Part B

Compute the magnitude of the radial acceleration of a point on its rim at the start.

Part C

Compute the magnitude of the resultant acceleration of a point on its rim at the start.

Part D

Compute the magnitude of the tangential acceleration of a point on its rim after it has turned through 60.0

Part E

Compute the magnitude of the radial acceleration of a point on its rim after it has turned through 60.0 .

Part F

Compute the magnitude of the resultant acceleration of a point on its rim after it has turned through 60.0 .

Part G

Compute the magnitude of the tangential acceleration of a point on its rim after it has turned through 120.0 .

Part H

Compute the magnitude of the radial acceleration of a point on its rim after it has turned through 120.0 .

Part I

Compute the magnitude of the resultant acceleration of a point on its rim after it has turned through 120.0 .

A) atan = r*alpha = 0.2*0.6 = 0.12 m/s^2

B) arad = r*wi^2 = 0.6*0 = 0 m/s^2

C) a = sqrt(0.12^2+0^2)= 0.12 m/s^2

D) 60 degrees = 60*pi/180 = 60*3.142/180 = 1.04 rad

theta = 1.04 rad

atan = r*alpha = 0.12 m/s^2

E) wf = sqrt(wi^2+(2*alpha*theta)) = sqrt(2*0.2*1.04) = 0.644 rad/s

arad = r*wf^2 = 0.6*0.644^2 = 0.248 m/s^2

F) a = sqrt(0.12^2+0.248^2) = 0.275 m/s^2

G) atan = r*alpha = 0.12m/s^2

H) theta = 120*3.142/180 = 2.09 rad

wf^2 = (2*alpha*theta) = 2*0.2*2.09 = 0.836

arad = r*wf^2 = 0.6*0.836 = 0.5016 m/s^2

I) resultant accelaration a = sqrt(atan^2+arad^2) = sqrt(0.12^2+0.5016^2) = 0.515 m/s^2

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