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)
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?
9 answers
step by step
I donot gate explained answer
Why did you say omega is 8πrad/SCE?
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
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
Knowledge
A, 8pi rad/s
B, 40rad.
B, 40rad.
to get answer
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
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
3 answers