- #1
bigtymer8700
- 40
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A light string is wrapped around the outer rim of a solid uniform cylinder of radius .375m that can rotate without friction about an axle through its center. A 3.00 kg stone is tied to the free end of the string When the system is released from rest, you determine that the stone reaches a speed of 3.50 m/s after having fallen 2.50m .
What is the mass of the cylinder?
Ive solved for:
F=M*a
3kg*9.8=29N
29N*2.5m=73.5J
i know its a solid cylinder so I=1/2 Mr^2 so i need to solve for the inertia. from here i got stuck i know i have to use
Kf=1/2mv^2+1/2Iw^2 i don't know where to go from here. do i solve for the angular velocity(w)? or Inertia?
i know v=rw so would w=v/r i could use 3.5m/s/.375m to get 9.3rad/s?
What is the mass of the cylinder?
Ive solved for:
F=M*a
3kg*9.8=29N
29N*2.5m=73.5J
i know its a solid cylinder so I=1/2 Mr^2 so i need to solve for the inertia. from here i got stuck i know i have to use
Kf=1/2mv^2+1/2Iw^2 i don't know where to go from here. do i solve for the angular velocity(w)? or Inertia?
i know v=rw so would w=v/r i could use 3.5m/s/.375m to get 9.3rad/s?