A spinning disk is rotating at a rate of 40 rad/s in the clockwise direction (vi

Question

A spinning disk is rotating at a rate of 40 rad/s in the clockwise direction (view from above) as shown in the figure above.
What is the direction of the angular velocity vector?
down
right
left
up
for this question the answer is down
Tries 0/2
If the disk is speeding up at a rate of 2 rad/s2, then what is the direction of the angular acceleration vector?
left
right
up
down
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Find the magnitude of the disk's angular velocity after 1 s.
rad/s
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Find the angle through which the disk has turned after 1 s in radians, degrees, and revolutions.
Angle through which wheel has turned (in radians) = radians
(in degrees) = degrees
(in revolutions) = revolutions
Tries 0/2
Find the magnitude of the velocity of a point located at r = 20 cm from the center of the disk after 1 s.
m/s
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Find the magnitude of the perpendicular acceleration of the point in the above part.
m/s2
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Find the magnitude of the parallel acceleration of the point in the above part.
m/s2
Tries 0/2
Find the total acceleration of the point in the above part.
m/s2

Explanation / Answer

to find the direction of angular velocity vector..
curl ur hand in the direction of motion..the thumb will point the direction of angular velocity vector..
so "down" is the answer...u already solved it..

speeding up implies acceleration should supportive in nature to angular velocity...hence same direction as angular velocity..."down"
f= 0+t
=> f = 40rad/s +2 rad/s =42rad/s ans
= 0t+(1/2)t2 = 41 rad ans

in degrees:
(in degree)= (in radian)(180/)
=2349.127 degree.
in revolution:
6.53 revolution


magnitude of velocity= f r= (42rad/s)0.2m= 8.4 m/s ans

magnitude of parpendicular acceleration= v2 /r= 352.8 m/s2


magnitude of parallel acceleration = 2rad/s2 (0.2)m=0.4m/s2

magnitude of total acceleration= vector sum of both perpendicular vectors=352.8002268 m/s2 ans

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