# Spinning disk torque?

spinning disk torque???

## Homework Statement

A uniform solid disk with radius 10 cm has mass 0.7 kg (moment of inertia I = ½MR2). A constant force 12 N is applied as shown. At the instant shown, the angular velocity of the disk is 40 radians/s in the -z direction (where +x is to the right, +y is up, and +z is out of the page, toward you). The length of the string d is 15 cm.

At this instant, what are the magnitude and direction of the angular momentum about the center of the disk?

What are the magnitude and direction of the torque on the disk, about the center of mass of the disk?

## The Attempt at a Solution

i know both of these have a -z direction. for finding the angular momentum i used
(MR^2/2)W (.7*.1^2/2) * (2(pi)/40) = 5.498e-4
then to find torque i used RF where F is tension force got this by (M*F) = (.7*12) = 8.4
then R*(8.4) = .1*8.4 = .84
what did i do wrong??

Do NOT multiply by the mass (.5kg). So, in short, multiply your final answer by 2 to get the correct one.

Do NOT multiply by the mass (.5kg). So, in short, multiply your final answer by 2 to get the correct one.

where did you get the mass of .5kg?? the mass in the problem is .7kg

i did that get the same answer as before. says im wrong

here is what i did

L(rot) = Iw
I = 1/2MR^2 so 1/2*.7*.1^2 = .0035
w = 2(pi)/T where T = 40 so w = .1570796
Iw = 5.497787e-4
do you see anything wrong with this??

alrite for part a) L(rot) = Iw
I=.5*.7*.1^2=.0035
L=.0035*40=.14

alrite for part a) L(rot) = Iw
I=.5*.7*.1^2=.0035
L=.0035*40=.14

OH.. ok
what i was doing was taking the 40rad/s
and using w = 2(pi)/T
so 2(pi)/40
then solving for that.

part b)

torque=abs(R)*abs(F)*sin(90)
=R*F*1=.1*12=1.2

should work

OH.. ok
what i was doing was taking the 40rad/s
and using w = 2(pi)/T
so 2(pi)/40
then solving for that.

Common mistake

for part d.) isn't it Lf-Li where Li is part a.) and Lf is I * omega with the given time?

for part d) you multiply torque calculated in part c) by the change in time