Back
to top
Sphere Under Gravity (circular motion)
Watch
Announcements
Page 1 of 1
Go to first unread
Skip to page:
I'm stuck on this question from Isaac Physics (Sphere Under Gravity).
A small sphere is attached to a fixed point by a string of length 𝑙=30cm , and whirls round in a vertical circle under the action of gravity at such speed that the tension in the string when the sphere is at its lowest point is three times the tension when the sphere is at its highest point.
Find the speed of the sphere at its highest point.
My working:
(F is the resultant/centripetal force)
At the bottom: F= 3T-mg
so mv2/r =3T-mg
At the top: F=T+mg
so mv2/r =T+mg
Rearranging the two equations and subbing one into the other to eliminate T:
mv2/r +mg=3mv2/r -3mg
Which cancels down and gets to v2/r =2g where r=0.3
so v=2.4248... ms-1
But that isn't the correct answer- why not?
A small sphere is attached to a fixed point by a string of length 𝑙=30cm , and whirls round in a vertical circle under the action of gravity at such speed that the tension in the string when the sphere is at its lowest point is three times the tension when the sphere is at its highest point.
Find the speed of the sphere at its highest point.
My working:
(F is the resultant/centripetal force)
At the bottom: F= 3T-mg
so mv2/r =3T-mg
At the top: F=T+mg
so mv2/r =T+mg
Rearranging the two equations and subbing one into the other to eliminate T:
mv2/r +mg=3mv2/r -3mg
Which cancels down and gets to v2/r =2g where r=0.3
so v=2.4248... ms-1
But that isn't the correct answer- why not?
0
reply
Report
#2
(Original post by BP_Tranquility)
I'm stuck on this question from Isaac Physics (Sphere Under Gravity).
A small sphere is attached to a fixed point by a string of length ������=30cm , and whirls round in a vertical circle under the action of gravity at such speed that the tension in the string when the sphere is at its lowest point is three times the tension when the sphere is at its highest point.
Find the speed of the sphere at its highest point.
My working:
(F is the resultant/centripetal force)
At the bottom: F= 3T-mg
so mv2/r =3T-mg
At the top: F=T+mg
so mv2/r =T+mg
Rearranging the two equations and subbing one into the other to eliminate T:
mv2/r +mg=3mv2/r -3mg
Which cancels down and gets to v2/r =2g where r=0.3
so v=2.4248... ms-1
But that isn't the correct answer- why not?
I'm stuck on this question from Isaac Physics (Sphere Under Gravity).
A small sphere is attached to a fixed point by a string of length ������=30cm , and whirls round in a vertical circle under the action of gravity at such speed that the tension in the string when the sphere is at its lowest point is three times the tension when the sphere is at its highest point.
Find the speed of the sphere at its highest point.
My working:
(F is the resultant/centripetal force)
At the bottom: F= 3T-mg
so mv2/r =3T-mg
At the top: F=T+mg
so mv2/r =T+mg
Rearranging the two equations and subbing one into the other to eliminate T:
mv2/r +mg=3mv2/r -3mg
Which cancels down and gets to v2/r =2g where r=0.3
so v=2.4248... ms-1
But that isn't the correct answer- why not?
What is the "correct" answer?
0
reply
https://isaacphysics.org/questions/sphere_under_gravity
I've entered that velocity but its saying its 'incorrect'. Also, just to make sure, the velocity at the top will be the same as the velocity at the bottom ?
0
reply
Report
#4
(Original post by BP_Tranquility)
Not sure what the correct answer is- this is the webpage the question is from:
https://isaacphysics.org/questions/sphere_under_gravity
I've entered that velocity but its saying its 'incorrect'. Also, just to make sure, the velocity at the top will be the same as the velocity at the bottom ?
Not sure what the correct answer is- this is the webpage the question is from:
https://isaacphysics.org/questions/sphere_under_gravity
I've entered that velocity but its saying its 'incorrect'. Also, just to make sure, the velocity at the top will be the same as the velocity at the bottom ?
The question doesn't really make it clear whether or not this mass moves at constant speed. It uses the phrase "at such speed", which could be interpreted as constant speed.
However, if you look at the hints it clearly shows that the speed is different. (v1 and v2). This would be the case normally, though many questions about swinging a mass on a string around in a vertical circle state that the speed is constant to make matters simpler.
It also states formulas for kinetic and potential energy.
The total energy remains constant but there is mgh potential potential energy at the top where h is 60cm, the diameter of the circle, and zero at the bottom.
You need to feed this into your calculation to find the difference in velocity at the top and bottom before feeding this into the centripetal force equations.
0
reply
(Original post by Stonebridge)
The question doesn't really make it clear whether or not this mass moves at constant speed. It uses the phrase "at such speed", which could be interpreted as constant speed.
However, if you look at the hints it clearly shows that the speed is different. (v1 and v2). This would be the case normally, though many questions about swinging a mass on a string around in a vertical circle state that the speed is constant to make matters simpler.
It also states formulas for kinetic and potential energy.
The total energy remains constant but there is mgh potential potential energy at the top where h is 60cm, the diameter of the circle, and zero at the bottom.
You need to feed this into your calculation to find the difference in velocity at the top and bottom before feeding this into the centripetal force equations.
The question doesn't really make it clear whether or not this mass moves at constant speed. It uses the phrase "at such speed", which could be interpreted as constant speed.
However, if you look at the hints it clearly shows that the speed is different. (v1 and v2). This would be the case normally, though many questions about swinging a mass on a string around in a vertical circle state that the speed is constant to make matters simpler.
It also states formulas for kinetic and potential energy.
The total energy remains constant but there is mgh potential potential energy at the top where h is 60cm, the diameter of the circle, and zero at the bottom.
You need to feed this into your calculation to find the difference in velocity at the top and bottom before feeding this into the centripetal force equations.
0
reply
Report
#6
(Original post by BP_Tranquility)
Hmm..It seems that if you use the conservation of energy so mgh=1/2 mv2 which reduces to v=root(2gh) and sub in h=0.6, then that gives the correct velocity of 3.43 ms-1. But surely that shouldn't work since that gives the velocity at the bottom, not the top?
Hmm..It seems that if you use the conservation of energy so mgh=1/2 mv2 which reduces to v=root(2gh) and sub in h=0.6, then that gives the correct velocity of 3.43 ms-1. But surely that shouldn't work since that gives the velocity at the bottom, not the top?
That gives the velocity at the bottom assuming the velocity at the top is zero. Which it clearly isn't. There is kinetic energy at both points. Just mgh more at the bottom.
0
reply
(Original post by Stonebridge)
That gives the velocity at the bottom assuming the velocity at the top is zero. Which it clearly isn't. There is kinetic energy at both points. Just mgh more at the bottom.
That gives the velocity at the bottom assuming the velocity at the top is zero. Which it clearly isn't. There is kinetic energy at both points. Just mgh more at the bottom.
0
reply
Report
#8
(Original post by BP_Tranquility)
So why did v=root2gh work? I would've thought I'd need to use the other info on Tension and centripetal force.
So why did v=root2gh work? I would've thought I'd need to use the other info on Tension and centripetal force.
It doesn't work. It gives an answer for the speed at the bottom if the speed is zero at the top, which, for some reason seems to happen to be the same as the correct answer.
It can't be correct because the tension would be zero at the top if the speed was zero.
0
reply
X
Page 1 of 1
Go to first unread
Skip to page:
Quick Reply
