Really confused i don't even understand the question?????

Question

A wheel 2.00 m in diameter lies in a vertical plane and rotates with a constant angular acceleration of 4.00 rad/s. The wheel starts at rest at t = 0, and the radius vector of a certain point P on the rim makes an angle of 57.3 with the horizontal at this time. At t = 2.00 s, find (a) the angular speed of the wheel, (b) the tangential speed and the total acceleration of the point P, and (c) the angular position of the point P

Answer 1

A wheel mounted on a wall is rotating. Every second, its velocity advances 4 radians/second, which is equivalent to 229.183 degrees/second. That is, if it is moving at 0 degrees/second at time 0, it will be moving at 229.183 degrees/second after 1 second, and it will be moving at 458.366 degrees/second after 2 seconds. This is actually the answer to part (a).

(a) The angular speed at t=2 seconds is 458.366 degrees/second, or 8 radians/second.

We assume that the wheel is rotating counter-clockwise. That is, if you look at it, a point at the top of the wheel will move to your left.

At t=0, we focus our eyes on a point on the rim of the wheel. That point is located 57.3 degrees above the horizontal. That is, if the wheel was a clock face, the point would be located around 1 o'clock. 1 o'clock is located at 60 degrees above the horizontal, so the point is actually located at where the hour hand would point at 1:05, which is just passed 1 o'clock. (keep in mind that 3 o'clock is located at 0 degrees and 12 o'clock is located at 90 degrees)

Now, at t=2 seconds, we already found that the wheel is rotating at 8 radians per second. Since the radius of the wheel is 2 meters, if the wheel moves 2*pi radians, the wheel moves 2*pi*r meters. Thus, there are r=2 meters per every radian. Thus,
the point at the edge of the wheel is traveling around the wheel at (8 radians per second)*(2 meters per radian) = 16 meters per second. Similarly, since the wheel is accelerating at 4.0 radians/second/second, the point at the edge of the wheel is accelerating along the circle that is the edge of the wheel at at (4.0 rad/s/s)*(2 meters/rad) = 8.0 meters/s/s.

However, there is also an acceleration normal to the edge of the circle pointing toward the center of the circle. That acceleration is thus a "centripetal" acceleration, and is given by v*v/r, where "v" is the tangential speed of the point. We know the tangential speed is 16 meters/second, so the centripetal acceleration is 128 m/s/s.

These two accelerations (tangential and normal/centripetal) are perpendicular to each other, so the TOTAL acceleration is the length of the hypotenuse of the right-triangle formed with each of them on a leg. That is, the total acceleration is the square root of ( (128 m/s/s)^2 + (8 m/s/s)^2 ), which is 128.25 m/s/s.

Thus, the answer to (b) is:

(b) The tangential speed is 16 meters/second. The tangential acceleration is 8 meters/s/s. The centripetal acceleration is 128 m/s/s. The TOTAL acceleration is 128.25 m/s/s.

Finally, you want to know the angle where P is after 2 seconds. Since everything started at rest, you know that (you can derive these with a little calculus; keep in mind that acceleration is change in velocity over time and velocity is change in position over time):

acceleration = 4 rad/s

velocity = acceleration*t = (4 rad/s)*t = 8 rad/s

DISPLACEMENT = 0.5*acceleration*t^2 = 0.5*(4 rad/s)*t^2 = 8 radians

So the angle of point P has moved 8 radians, which is equal to 458.366 degrees. Since the point P has started at 57.3 degrees, then you just add 57.3 degrees to this number and you get 515.666 degrees. Now, there are only 360 degrees in a circle, so this means that the point has completed one revolution and has gone 155.666 degrees into the next circle. Thus, your answer to (c) is:

(c) The angular position of the point P is 155.666 degrees.

I hope that helps.

Answer 2

You're right, it makes no sense the way you copied it here.

"rad/s" is not an acceleration unit, it is an angular speed unit. So let's assume you meant alpha = 4 rad/sec^2. If that's the case, then w = w0 + alpha t; where w is the final angular velocity (rad/sec) after accelerating for t = 2 seconds and starting with initial angular velocity w0 at t = 0.

Thus, w = w0 + alpha t = 0 + 4 rad/sec^2 2 sec = 8 rad/sec for answer (a).

The tangential speed (velocity) is just the radius of the wheel (r) times the angular speed (w); so that is v = r w = 2 m X 8 rad/sec = 16 m/sec. I have no clue what "total acceleration" means, if that means tangential acceleration (the acceleration in the v direction); then that is a = r alpha = 2 m X 4 rad/sec^2 = 8 m/sec^2 presuming the angular acceleration is constant.

For (c) the angular position is P = P0 + avgw t; where P0 = 1 rad (57.3 deg with the horizontal) the initial starting point for the problem and avgw = 1/2 (w0 + w) = 1/2 (0 + 8 rad/sec) = 4 rad/sec, which is the average angular velocity over the two seconds of travel following t = 0. Thus the final position after two seconds of acceleration is P = P0 + avgw t = 1 rad + 4 rad/sec X 2 sec = 9 rad. Or in degrees, 9 rad X 57.3 deg/rad (a bit over 515 deg) measured from when the point (P) was horizontal at t = 0. So P on your wheel will have circled more than once around from the starting horizontal position.

Answer 3

The above answers are correct except for b) the TOTAL
acceleration of point P.

Since P moves in a circle, there is an inward acceleration
it has in ADDITION to the tangential acceleration.

The inward centripetal force is mv^2/r, so the inward acceleration is v^2/r. The v here is the tangenial speed which is
16 m/s.

You then have to add to this the tangential acceleration of
8 m/s/s. Remember this is vector addition!!

Answer 4

We can very well apply the kinematic
equations in circular rotation as in this problem.
a)angular speed after 2sec
v=u+at
u=0
a=4rad/sec
t=2sec
v=4*2=8rad./sec
b)Tangential speed
=radius*angular velocity
=2m*8rad/sec
=16m/sec
c)angular position
=u*t+1/2*a*t^2 [u=0]
=1/2*4*2^2
=8radians
Pi*rad.=180Deg
rad=180/Pi
8rad=8*180*7/22Deg
=360*28/22Deg
[Division by 360 will not alter
the position of radius vector]
=14/11 deg
=1deg 16min 21sec
the position of radius vector
would be
57deg20min.+1deg 16min. 21sec
=58Deg 36min 21sec