Question

Point **P**is on the rim of a large
**Spinning Disk** of mass 10.0 kg and radius 2.58 m.
At time t=0.00 s the disk has an angular velocity of 4.00 rad/s and
rotates counterclockwise about its center **O**, and
**P**is on the x-axis. A net applied CW torque of 20.0
m-N causes the wheel to undergo a uniform angular acceleration. The
magnitude of the total acceleration (m/s^{2}) at point
**P**when t = 15.0 s is? The angular velocity of the
disk at 15.0 s is? The rotational kinetic energy of the disk at
15.0 s is?

Answer #1

**moment of inertia I = (1/2)*M*R^2**

**I = (1/2)*10*2.58^2 = 33.282 kg m^2**

**torque = -20.0 Nm**

**torque = I*alpha**

**angualr acceleration alpha = -20/33.282 = -0.6
rad/s^2**

**at tiemt = 15 s**

**angular speed w = wo + alpha*t**

**w = +4 - (0.6*15) = -5 rad/s**

**radial accelration arad = R*w^2 = 2.58*5^2 = 64.5
m/s^2**

**tangentaila acceleration atan = R*alpha = -2.58*0.6 =
-1.548 m/s^2**

**total acceleration atot = sqrt(atan^2
+arad^2)**

**atot = sqrt(1.548^2+64.5^2) = 64.5 m/s^2**

**-----------------------------**

**anguilar velcoity w = 5 rad/s clockwise**

**---------------------**

**kinetic energy K = (1/2)*I*w^2 = (1/2)*33.282*5^2 = 416
J**

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