help me out here!!!
#11
12-18-2007, 01:30 PM
RE: help me out here!!!
it's always the Butler - I don't like that Col Mustard - I mean of all the charaters he's the only trained assasin, right?!?!?
#13
12-18-2007, 03:57 PM
RE: help me out here!!!
You're still missing a time component. Velocity is a rate of change of position. Speed and direction are both required to figure out velocity. Since speed is defined as a change in distance per unit of time, the problem can't be completed as stated. Anyways, I have much more important things to get to, such as how to balance a compressor off of a 60-year-old turbine engine.
#14
12-18-2007, 04:16 PM
RE: help me out here!!!
ORIGINAL: scotts custom
alright so im bored and im "studying" for finals, so ill let you see if you can figure out any of this stuff that i have to learn in a few hours time. I got a lot more, so if you happen to know this stuff and want to work your brain, dont be scared to ask..
lets see...
A .2kg ball is tied to a rope and swung around in the air in a circular path (like a lasso if you will) and its radius of rotation is .75 meters,
1) what is its angular velocity, 2) tangential velocity and 3) the tension force on the rope?
dont have too much fun!!!
alright so im bored and im "studying" for finals, so ill let you see if you can figure out any of this stuff that i have to learn in a few hours time. I got a lot more, so if you happen to know this stuff and want to work your brain, dont be scared to ask..
lets see...
A .2kg ball is tied to a rope and swung around in the air in a circular path (like a lasso if you will) and its radius of rotation is .75 meters,
1) what is its angular velocity, 2) tangential velocity and 3) the tension force on the rope?
dont have too much fun!!!
There should be one that allows you to disregard time. What class is this for and are you using the angular quantities or using the [T.C.] directional trick to bypass needless calculation?
#17
12-18-2007, 04:44 PM
Senior Administrator
Join Date: Feb 2005
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Posts: 13,574
RE: help me out here!!!
^^that assumes the rope is taut, which isn't stated. 3/4 meter could be with teh weight only circling at 1/2 speed, or full speed. We don't know. The radius is given, but not teh lenght of the rope.
#18
12-18-2007, 05:04 PM
RE: help me out here!!!
ORIGINAL: AWDaholic
^^that assumes the rope is taut, which isn't stated. 3/4 meter could be with teh weight only circling at 1/2 speed, or full speed. We don't know. The radius is given, but not teh lenght of the rope.
^^that assumes the rope is taut, which isn't stated. 3/4 meter could be with teh weight only circling at 1/2 speed, or full speed. We don't know. The radius is given, but not teh lenght of the rope.
#19
12-19-2007, 02:46 PM
RE: help me out here!!!
I'd like to know the answer to this problem. I do feel you need a time component.
The guy who said "there should be an equation that doesn't need time" or whatever... I think you are thinking of the constant acceleration equations. Although... maybe they can be applied to rotation. Sometimes...
The guy who said "there should be an equation that doesn't need time" or whatever... I think you are thinking of the constant acceleration equations. Although... maybe they can be applied to rotation. Sometimes...
#20
12-19-2007, 02:56 PM
RE: help me out here!!!
ORIGINAL: AWDaholic
^^that assumes the rope is taut, which isn't stated. 3/4 meter could be with teh weight only circling at 1/2 speed, or full speed. We don't know. The radius is given, but not teh lenght of the rope.
^^that assumes the rope is taut, which isn't stated. 3/4 meter could be with teh weight only circling at 1/2 speed, or full speed. We don't know. The radius is given, but not teh lenght of the rope.
You also assume ropes are massless, you can't push with a rope, and that they don't flex or twist.