A 52.4-cm diameter disk rotates with a constant angular acceleration of 2.50 rad

Question

A 52.4-cm diameter disk rotates with a constant angular acceleration of 2.50 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.

(c) Find the position of P (in degrees, with respect to the positive x-axis) at t = 2.30s.
__________°

Explanation / Answer

(a)   The angular speed of the wheel at t = 2.30 s,

= *t

= (2.5)*(2.3)  = 5.75 rad/s

(b)   The linear velocity and tangential acceleration of P at t = 2.30 s,

v = *r , where r = d/2 = 52.4 / 2 = 26.2 cm

= (5.75)*(26.2)  = 150.65 cm/s  = 1.51 m/s

a = *r

= (2.5)*( 26.2) = 65.5 cm/s^2 = 0.655 m/s^2

(c) The position of P (in degrees, with respect to the positive x-axis) at t = 2.30s,

f = i + (1/2)**t^2

In this case, i = 57.3°, which is 1 rad, so:

f = 1 + (1/2) *(2.5)*(2.3)^2

= 1 + 5.75  = 6.75 rad = 386.78°

=> f = 26.78° with respect to the positive x-axis.

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