# Rotational Motion and speed of a disc

1. Jun 22, 2012

### nbroyle1

A 36.5-cm diameter disk rotates with a constant angular acceleration of 2.00 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.
(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.

(a) to solve for angular speed W=at or W=(2)(2.3) and got 4.6m/s

(b) first I calculated the radius by multiplying 36.5*.01 then dividing by 2 and got .1825.

next I plugged it into V=(.1825)(4.6) to solve for linear velocity and got .8395.

To get tangential acceleration I used A=(.1825)(2) and got .365.

(c) θ=1+.5(2)(2.3)^2 and got 6.29 radians which I converted to degrees and got 360.39 and subtracted 360 to get position with respect to positive x-axis.

Not sure If I did part this right can I get some verification??

2. Jun 22, 2012

### Zubeen

Well ..... why u added 1 to θ in the (c) part ?
the formula simply says-
θ=ωot + αt2
and ωo is initial rotational velocity.... and since disk was at rest at t=0 so ωo = 0
rest all is correct.

3. Jun 22, 2012

### nbroyle1

The reason I did that in part c was because I subtracted θo to the right side of the equation. I thought the initial angular displacement would be 1 since it is one radian.

4. Jun 22, 2012

### nbroyle1

I meant I added it to the right side of the equation oops

5. Jun 22, 2012

### Zubeen

...... sorry ..... yes .... you are correct .
even i meant Δθ by θ.

6. Jun 22, 2012

### nbroyle1

ok thanks

7. Jun 23, 2012

### Zubeen

yes .... but the direction of angular acceleration will make a difference, if its clockwise or anticlockwise, the final angle with the x-axis would be different in the two cases.