wf=alpha*time
where alpha=torque/momentinertia
= 5.5*.36/(1/2 *4*.18^2)
wf will be in rad /sec. to get rpm
RPM=wf/2pi * 60/1=30wf/PI
A 4.0 kg, 36-cm-diameter metal disk, initially at rest, can rotate on an axle along its axis. A steady 5.5 N tangential force is applied to the edge of the disk.
What is the disk's angular velocity, in rpm, 4.0 s later?
3 answers
Here is the answer I got for anyone wondering:
T - Torque
a - angular acceleration
F - Tangential Force
r - radius
I - Inertia
w - angular velocity
t - time
M - mass
Torque has two different equations associated with it, and we can set these equations equal to solve the problem
T=aI , T=Frsin(theta) --> aI=Frsin(theta) --> a=(Frsin[theta])/I
I (inertia)=(0.5)Mr^2 since we can treat this as a solid cylinder and the axis of rotation going through the center of mass
So, a=(Frsin[thea])/(0.5Mr^2)
Finally use rotational equations to solves.
w=at --> w=t*(Frsin[theta])/(0.5mr^2)
w=4s*(5.5N*0.18m*sin90)/(0.5*4kg*0.18m^2)=61.11rad/s
convert to rpm and that is 61.11rad/s*60s/(2pi rad)=583.57rpm
So, after 4 seconds, the angular velocity is 580rpm.
T - Torque
a - angular acceleration
F - Tangential Force
r - radius
I - Inertia
w - angular velocity
t - time
M - mass
Torque has two different equations associated with it, and we can set these equations equal to solve the problem
T=aI , T=Frsin(theta) --> aI=Frsin(theta) --> a=(Frsin[theta])/I
I (inertia)=(0.5)Mr^2 since we can treat this as a solid cylinder and the axis of rotation going through the center of mass
So, a=(Frsin[thea])/(0.5Mr^2)
Finally use rotational equations to solves.
w=at --> w=t*(Frsin[theta])/(0.5mr^2)
w=4s*(5.5N*0.18m*sin90)/(0.5*4kg*0.18m^2)=61.11rad/s
convert to rpm and that is 61.11rad/s*60s/(2pi rad)=583.57rpm
So, after 4 seconds, the angular velocity is 580rpm.
The final conversion to rpm is incorrect