Question

Point Pis 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 Pis 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/s²) at point Pwhen 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

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anguilar velcoity w = 5 rad/s clockwise

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