An electric turntable 0.710 m in diameter is rotating about a fixed axis with an initial angular velocity of 0.230 rev/s. The angular acceleration is 0.887 rev/s^2.
A. Compute the angular velocity after a time of 0.192 s.
B. Through how many revolutions has the blade turned in this time interval?
C. What is the tangential speed of a point on the tip of the blade at time t = 0.192 s?
D. What is the magnitude of the resultant acceleration of a point on the tip of the blade at time t = 0.192 s?
Please help, thanks!
A. Compute the angular velocity after a time of 0.192 s.
B. Through how many revolutions has the blade turned in this time interval?
C. What is the tangential speed of a point on the tip of the blade at time t = 0.192 s?
D. What is the magnitude of the resultant acceleration of a point on the tip of the blade at time t = 0.192 s?
Please help, thanks!
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A) The angular velocity w = w0 + a*t where w0=0.23*2*Pi rad./s
also where 'a' = 0.887*2*Pi rad./s^2
Thus at t=0.192, w(0.192) = 0.23*2*Pi + 0.887*2*Pi*0.192
=2.5 rad/s
B) Let angular displacement be 'd'
d(t) = integral {w0 + a*t}dt between 0 and 0.192 s
=w0*t+a*(t^2)/2+c
d(0)=0=c
d(0.192)= 0.23*2*Pi*0.192 + 0.887*2*Pi/2*0.192^2
=0.38 radians. This is 0.38/(2*Pi)=0.06 revolutions.
C) tangential speed = w*r = 2.5*0.71=1.775 m/s
D) resultant accn = sqrt((centripetal accn)^2 + (tangential accn)^2)
=sqrt((w^2*r)^2 + (a*r)^2)
=sqrt((2.5^2*0.71)^2 + (0.887*2*Pi*0.71)^2)
=5.95 m/s^2
also where 'a' = 0.887*2*Pi rad./s^2
Thus at t=0.192, w(0.192) = 0.23*2*Pi + 0.887*2*Pi*0.192
=2.5 rad/s
B) Let angular displacement be 'd'
d(t) = integral {w0 + a*t}dt between 0 and 0.192 s
=w0*t+a*(t^2)/2+c
d(0)=0=c
d(0.192)= 0.23*2*Pi*0.192 + 0.887*2*Pi/2*0.192^2
=0.38 radians. This is 0.38/(2*Pi)=0.06 revolutions.
C) tangential speed = w*r = 2.5*0.71=1.775 m/s
D) resultant accn = sqrt((centripetal accn)^2 + (tangential accn)^2)
=sqrt((w^2*r)^2 + (a*r)^2)
=sqrt((2.5^2*0.71)^2 + (0.887*2*Pi*0.71)^2)
=5.95 m/s^2