Recent content by tsnikpoh11
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What Distance Does the Bucket Cover in 3.5 Seconds?
[SOLVED] How far does the bucket fall? Homework Statement A cylindrical pulley with a mass of M = 5.7 kg and a radius of r = 0.66 m is used to lower a bucket with a mass of m = 2 kg into a well. The bucket starts from rest and falls for 3.5 s. The acceleration of gravity is 9:8 m/s^2 ...- tsnikpoh11
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- Fall
- Replies: 1
- Forum: Introductory Physics Homework Help
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What is the Moment of Inertia for a Uniform Disk with a Falling Object?
Homework Statement A uniform disk of radius 0.12 m is mounted on a frictionless, horizontal axis. A light cord wrapped around the disk supports a 1 kg object, as shown in the figure from rest the object falls with a downward acceleration of 2.3 m/s^2.The acceleration of gravity is 9.8 m/s^2...- tsnikpoh11
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- Disk Inertia Moment Moment of inertia
- Replies: 1
- Forum: Introductory Physics Homework Help
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Magnitude of the angular acceleration about the pivot point?
Awesome! Thank you so much.- tsnikpoh11
- Post #7
- Forum: Introductory Physics Homework Help
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Magnitude of the angular acceleration about the pivot point?
Ohhhhh, ok so its 8.628+-gravity/5.184 ? t/I? Would the force down due to gravity be mg(l/2)= force of gravity? if so do I subtract that from the 8.628 force up? so the force of gravity is 31.704 and the torque up is 8.628? Then would I have to convert that answer to radians?- tsnikpoh11
- Post #5
- Forum: Introductory Physics Homework Help
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Magnitude of the angular acceleration about the pivot point?
So Net torque would = 2.4 *4(sin64)=8.628? or the net force is 9.81 down - 8.832 up? = -.968 Net force? How do I account for gravity? So the moment of Inertia would be (ML^2)/3 which is I=(2.7*(2.4)^2)/3 = 5.184, but how does it work into Newtons second law, and how does it help me...- tsnikpoh11
- Post #3
- Forum: Introductory Physics Homework Help
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Magnitude of the angular acceleration about the pivot point?
[SOLVED] Magnitude of the angular acceleration about the pivot point? Homework Statement A uniform horizontal rod of mass 2.7 kg and length 2.4 m is free to pivot about one end as shown. The moment of inertia of the rod about an axis perpendicular to the rod and through the center of mass...- tsnikpoh11
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- Acceleration Angular Angular acceleration Magnitude Pivot Point
- Replies: 6
- Forum: Introductory Physics Homework Help