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EK 1001 Physics #203
Started by Damia
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well, ~28 is what the squareroot of that answer. maybe if you post the problem it can make more sense?
set the maximum tension equal to centripetal force and solve for v:
T = mv^2/R
1600 N = (1.5 kg)(v^2)/(1.5 m)
v^2 = 1600
v = (1600)^(1/2) = 40 m/sec
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WOW. THIS IS JUST WRONG
this is why I hate review books. First off the maximum tension held is always at the top of the loop. In essence the max velocity is determined by the forces acting at the top. At the top the two forces acting are inward-weight and tension. The only countering force is the objects velocity which wants to move tangentially (and not circular as caused by the string). Hence this is represented by centripetal force.
Thus: Weight plus tension =mv^2/r. So tension equals mv^2/r - weight. As you can see the weight term causes the velocity to be HIGHER than if you were consdering the tension alone.
Likewise. When one asks what the minimum velocity is for the object to spin in a circle the same equation is used above (since the slowest the object can go is at the top-if it were slower than that the tension would go slack and the object would fall down). In any case the equation above is used but tension is zero this time and mv^2/r = weight
this is why people should read their textbooks!!!
this is why I hate review books. First off the maximum tension held is always at the top of the loop. In essence the max velocity is determined by the forces acting at the top. At the top the two forces acting are inward-weight and tension. The only countering force is the objects velocity which wants to move tangentially (and not circular as caused by the string). Hence this is represented by centripetal force.
Thus: Weight plus tension =mv^2/r. So tension equals mv^2/r - weight. As you can see the weight term causes the velocity to be HIGHER than if you were consdering the tension alone.
Likewise. When one asks what the minimum velocity is for the object to spin in a circle the same equation is used above (since the slowest the object can go is at the top-if it were slower than that the tension would go slack and the object would fall down). In any case the equation above is used but tension is zero this time and mv^2/r = weight
this is why people should read their textbooks!!!
In the given example the ball is swinging HORIZONTALLY around the boy's head and thus gravity and weight are irrelevant because they act perpendicular to the plane of the ball's motion.
And in the case where the ball was swinging vertically, then the max tension would be at the bottom of the swing with:
Tmax = m(v^2/r + g)
while minimum tension would be at the top of the swing:
Tmin = m(v^2/r - g)
and if v^2/r < g then the ball would fail to swing in a perfect circle
nowhere did his question say it was horizontal. If so then that should be clearer. Second I was referring to speed-not tension. Max speed is at top. Third you are wrong about minimum tension for it to swing-it is zero at the top. The tension cannot be lower than this. Since minimum tension is zero at top for it to spin(vertical spin) then by default the lowest tensioncannot be zero at bottom-there tension must equal weight and cetripetal
When it said "around his head," it is common sense that it is horizontal.
