A pully, with a rotational inertia of 1.0 x 10^-3kg.m^2 about its axle and a radius of 10 cm,?

Question

is acted on by a force applied tangentially at its rim. The force magnitude varies in time as F = 0.50t + 0.30t^2, with F in newtons and t in seconds. The pulley in initially at rest. At t = 3.0s, what are (a) its angular acceleration and (b) its angular speed?

1057

Answer 1

a)
The force at 3 s is
F(3) = 0.50(3)+ 0.30(3)^2 = 4.2 N

Torque = I (alpha)
alpha = Torque / I = rF/ l
alpha= (4.2)(0.1) / (1.0 x 10^-3) = 420 rad/s/s


b)

At t=3 s
dw/dt=alpha
dw=(alpha)dt= (rF/I)dt=(r/I)(0.50t + 0.30t^2)dt
w = (r/I )(9/4 +27)= (0.1/0.001)(4.95)= 495 rad/s

Answer 2

I alpha = F r; where F(t) = 1/2 t + .3 t^2 and r = .1 m. Where the angular acceleration you are looking for is alpha(t) = F(t) r/I; where I = 10^-3 kg m^2.

(a) alpha(t) = (1/2 t + .3 t^2) r/I at t = 3 sec. Everything on the RHS is given; so you can do the math.

(b) w(t) = INT(alpha(t)), the angular velocity is the integration of alpha(t) over t; thus, w(t) = (1/4 t^2 + .1 t^3) r/l at t = 3 sec. You can do the math.

The physics is this. Rotational inertia (I) is just a math concept that is the analog to mass (m) in linear equations for force and momentum. Thus we have I alpha = F r, which is the analog to m a = F; where a is linear acceleration. And we also have I w = m v r, which is the analog to m v = p; where v is the linear velocity, w is angular velocity, and p is the linear momentum.

To see how useful I is, note that KE(A) = 1/2 I w^2, which is angular kinetic energy. The linear analog is KE(L) = 1/2 m v^2; where one can see that the m takes the place of I and v takes the place of w in the angular version.

If alpha were a constant, we could use the angular equivalents to the SUVAT equations. But it is not. So we need to integrate the alpha(t) equation to find w(t) the angular velocity as a function of t.

Answer 3

At t = 3 s , the force F = 0.5*3 + 0.3*9 = 4.2 N
The torque T = F r = 4.2*0,1 = 0.42 Nm
T = I alpha ---> alpha(3) = 0.42/1*10^-3 = 420 rad/s^2
(b) dw/dt = (r/I) (0.5 t + 0.3 t^2)
w(3) = 100(0.25*9 + 0.1*27) = 495 rad/s