Asked by Nevaeh
a stop over 3 seconds. What is the angular acceleration? B. How many radians did the disk turn while stopping ? C. how many revolutions?
R=25 cm= 0.25 m
F=30 rpm w0=2pif=2pi(30)=60pi rad/min
W0=60pi rad/60s=pi rad/s
W=at+w0 0=a(3)+pi a= -pi/3
Theta=1/2at^2+w0t=1/2(-pi/3)(3)^2+pi(3)=-3pi/2+3pi=3pi/2 rad
Theta=2pin=n=theta/2pi=3pi/2/2pi=0.75 rev= 3/4 rev
Is this correct or not if it's not correct me.
R=25 cm= 0.25 m
F=30 rpm w0=2pif=2pi(30)=60pi rad/min
W0=60pi rad/60s=pi rad/s
W=at+w0 0=a(3)+pi a= -pi/3
Theta=1/2at^2+w0t=1/2(-pi/3)(3)^2+pi(3)=-3pi/2+3pi=3pi/2 rad
Theta=2pin=n=theta/2pi=3pi/2/2pi=0.75 rev= 3/4 rev
Is this correct or not if it's not correct me.
Answers
Answered by
Anonymous
R = .25 meter
omega at start= wi = 30 * 2 pi /60 = pi rad/s
w = wi + a t
0 = pi + a (3)
a = - pi/3 {agree}
theta = wi t + (1/2) a t^2 = 3 pi - (pi/6)(9) = pi ( 3-1.5) = 1.5 pi
yes that is 3/4 rev
omega at start= wi = 30 * 2 pi /60 = pi rad/s
w = wi + a t
0 = pi + a (3)
a = - pi/3 {agree}
theta = wi t + (1/2) a t^2 = 3 pi - (pi/6)(9) = pi ( 3-1.5) = 1.5 pi
yes that is 3/4 rev
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