A steel rod of 500g and 30 cm long spins at 300 RPM,the rod pivots around the center.
a) Find angular momentum
Moment of Inertia(I)=mk^2
I=0.5kgX(0.075m)^2
I=0.0028125kg*m^2
To find angular momentum:
w=300rpmX2pi/60sec
w=31.42 rads/s
Ang. Momentum=Iw
A.M.=0.0028125X31.42
A.M.=0.08837kg*m^2/s
A.M.=0.08837Nm/s or J/s (Ans.)
b)Find the average torque needed to stop the rod in 2 seconds.
Deceleration needed to stop the rod in 2sec=w2-w1/time
Dec.=0-31.42 rads/s/2sec
Dec.=-15.7rads/s^2
Then:Torque=IXdec.
T=0.0028125kg*m62X15.7rads/sec^2
Torque=0.044Nm (Ans.)
Are my calculations correct?
3 answers
I thought the moment of inertia would be 1/12 ml^2
What do you mean by: 1/12ml^2
k^2=radius of gyration
k^2=radius of gyration
I am using l as the length of the rod. (.3m)
Recheck your formula.
Recheck your formula.
1 answer