Thread: Torque problem

Er = la = l(w^2 - w0^2)/2*theta

this is for Torque i believe

A CD has a mass of 17 g and a radius of 6.0 cm. When inserted into a player, the CD starts from rest and accelerates to an angular velocity of 18 rad/s in 0.70 s. Assuming the CD is a uniform solid disk, determine the net torque acting on it.
in N·m

i could convert 17 to kg if i needed to.

i know the angular velocity starts at 0.

and the final angular velocity is 18.

if i use angular kinematics i can use theta f = theta i + 1/2(wi + wf) * t

to get theta f = 6.3
i believe.

so i'm not sure why this does not work: radius = 6 ... 6 x 2 = 12 cm (diameter)

so then, 12(18^2 + 0^2)/ 2*(6.3)

I got 308.57 N*m and that is not correct.

Originally Posted by rcmango

A CD has a mass of 17 g and a radius of 6.0 cm. When inserted into a player, the CD starts from rest and accelerates to an angular velocity of 18 rad/s in 0.70 s. Assuming the CD is a uniform solid disk, determine the net torque acting on it.

I'm not sure where you got your method, nor your last equation.

$\displaystyle \sum \tau = I \alpha$
where
$\displaystyle I = \frac{1}{2}mr^2$

So as you said:
$\displaystyle \omega = \omega _0 + \alpha t$

$\displaystyle 18 = 0 + \alpha (0.70) \implies \alpha = 25.7143~rad/s^2$

So
$\displaystyle \sum \tau = \frac{1}{2}(0.017~kg)(0.06~m)^2(25.7143~rad/s^2) = 7.869 \times 10^{-4}~Nm~rad$

(I personally prefer the Nm rad unit for torque. Most others don't.)

-Dan