A 47.3-cm diameter disk rotates with a constant angular acceleration of 2.70 rad

Question

A 47.3-cm diameter disk rotates with a constant angular acceleration of 2.70 rad/s2. It starts from rest at t = 0, and a line drawn from the center of the disk to a point P on the rim of the disk makes an angle of 57.3° with the positive x-axis at this time. (a) Find the angular speed of the wheel at t = 2.30 s. rad/s (b) Find the linear velocity and tangential acceleration of P at t = 2.30 s. linear velocity m/s tangential acceleration m/s2 (c) Find the position of P (in degrees, with respect to the positive x-axis) at t = 2.30s. °

Explanation / Answer

Using velocity notations in place of angular velocity for ease of use. Please take note of it.

at t=0 sec, a=2.7, s= 57.3

part a

v(t=2.30)= u+at

= 0+ 2.7*2.3

= 6.21 rad/s

part b

linear velocity= (angular velocity)*radius of disk

= 6.21*(47.3/2)*10-2

= 1.46 m/s

tangential acceleration= (angular acceleration)*radius of disk

= 2.7*(47.3/2)*10-2

        = 0.64 m/s2

part c

change in angle = ut+0.5*a*t^2

= [0*2.3+0.5*2.7*2.3^2]

= 7.14 radians

Now. 7.14 rad= 204.69 degree

So, final angle= 57.3+204.69 = 261.95 degree with positive x axis

= (-) 98.05 degree with positive x axis

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