Asked by Emanuel silver(Ag)

A spinning top has a diameter of 10 cm. A point on the outer rim of the top moves through an angle of 8¶ radians each second. A) what is the angular velocity of the point? B) what is the distance moved by the point in 5 seconds? C) what is the velocity of the point? D) what is the acceleration of the point?
Answered by Damon
R = 0.05 meter
A. You already said omega = 8 pi radians/second

C. omega R = v = 8 pi (0.05)

B. theta = angle moved = (8 pi)(5)
so distance = R Theta = (0.05)(8 pi)(5)

D. Ac = R omega^2 = 0.05 (64 pi^2)
Answered by ela
step by step
Answered by Rezika
I donot gate explained answer
Answered by Israel Bekele
Why did you say omega is 8πrad/SCE?
Answered by Hanan neb
a .8 rad / sec
b. ፀ = tw = 5s x 8pi rad/s = 40pi rad
c. V= r w= 0.05 x 8 = 0.4
d. a= v squre/r =0.16/ 0.05 = 3.2
Answered by Mohamed
Knowledge
Answered by Abiyot degu
A, 8pi rad/s
B, 40rad.
Answered by Mesfin
to get answer
Answered by Nuresselam
a,8¶r per s=4r/s b/s 1revo =360°=2¶r so
W=4rad/s
b,teta= wt=4rad/s*5s=20rad
C,v=wr=4rad/s*0.05m= 0.2m/s
d,a=w^2*r=16*0.05=0.8m/s^2
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