- #1
Ishaan S
- 19
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[Note from mentor: this thread was originally posted in one of the non-homework forums, therefore it does not use the standard homework template.]
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Hi,
I am self-studying ap physics, and as I have no teacher to guide me, am seeking help online. I have a problem that I have been trying to solve for about an hour now. I had thought I had the answer, but when I saw the answer, it was very close, but it was the negative version of what I had. Here's the problem:
The sides of a cone make an angle φ with the vertical. A small mass m is placed on the inside of the cone and the cone, with its point down, is revolved at an angular velocity ω (radians/sec) about its symmetry axis. If the coefficient of static friction is μ, at what positions on the cone can the mass be placed without sliding on the cone? (Give max and min distances, r, from the axis).
The answer the book gives is:
rmax=(g/ω2)((1+μtanφ)/(tanφ+μ))
rmin=(g/ω2)((1-μtanφ)/(tanφ+μ))
My attempt at a solution:
I broke the centripetal force into its components: One component is inwardly perpendicular to the surface
mω2rcosφ
The other is pointing toward the tip(downward parallel to the surface). This component is
mω2rsinφ
The gravity perpendicular to the surface is outward from the cone
mgsinφ
The parallel component is toward the tip of the cone parallel to the surface of the cone
mgcosφ
The friction would be μ*FN≥Ff.
The normal force here is
FN=mω2rcosφ - mgsinφ
I substitute the normal force into the friction inequality.
I now applied the F=ma formula in the direction parallel to the surface, using the direction of the tip of the cone as positive. a=0, so it's just F=0.
mω2rsinφ + mgcosφ - Ff = 0
I substituted in the max static friction and solved for r.
I got
r = (g/ω2)((1 + μtanφ)/(μ - tanμ))
This is off from the r max by a factor of -1. I have no idea why. The radius shouldn't be negative. Could someone please help me with that?
Also, if I make the friction negative and solve for r, I would get
r = (g/ω2)((μtanφ - 1)/(μ + tanφ)
This is also off by a factor of -1 for the minimum r.
Something I suspect is messing up my answer is that I have the centripetal force acting inward to the cone, perpendicular to the axis of revolution. If this were so, I would think that the object wouldn't fly out because all of the components are pointing toward the bottom tip. Could someone please provide a free-body diagram?
Help would be greatly appreciated. I want to thoroughly understand this, so it would be great if you explained what I did wrong and didn't just tell me what to do.
Thanks
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Hi,
I am self-studying ap physics, and as I have no teacher to guide me, am seeking help online. I have a problem that I have been trying to solve for about an hour now. I had thought I had the answer, but when I saw the answer, it was very close, but it was the negative version of what I had. Here's the problem:
The sides of a cone make an angle φ with the vertical. A small mass m is placed on the inside of the cone and the cone, with its point down, is revolved at an angular velocity ω (radians/sec) about its symmetry axis. If the coefficient of static friction is μ, at what positions on the cone can the mass be placed without sliding on the cone? (Give max and min distances, r, from the axis).
The answer the book gives is:
rmax=(g/ω2)((1+μtanφ)/(tanφ+μ))
rmin=(g/ω2)((1-μtanφ)/(tanφ+μ))
My attempt at a solution:
I broke the centripetal force into its components: One component is inwardly perpendicular to the surface
mω2rcosφ
The other is pointing toward the tip(downward parallel to the surface). This component is
mω2rsinφ
The gravity perpendicular to the surface is outward from the cone
mgsinφ
The parallel component is toward the tip of the cone parallel to the surface of the cone
mgcosφ
The friction would be μ*FN≥Ff.
The normal force here is
FN=mω2rcosφ - mgsinφ
I substitute the normal force into the friction inequality.
I now applied the F=ma formula in the direction parallel to the surface, using the direction of the tip of the cone as positive. a=0, so it's just F=0.
mω2rsinφ + mgcosφ - Ff = 0
I substituted in the max static friction and solved for r.
I got
r = (g/ω2)((1 + μtanφ)/(μ - tanμ))
This is off from the r max by a factor of -1. I have no idea why. The radius shouldn't be negative. Could someone please help me with that?
Also, if I make the friction negative and solve for r, I would get
r = (g/ω2)((μtanφ - 1)/(μ + tanφ)
This is also off by a factor of -1 for the minimum r.
Something I suspect is messing up my answer is that I have the centripetal force acting inward to the cone, perpendicular to the axis of revolution. If this were so, I would think that the object wouldn't fly out because all of the components are pointing toward the bottom tip. Could someone please provide a free-body diagram?
Help would be greatly appreciated. I want to thoroughly understand this, so it would be great if you explained what I did wrong and didn't just tell me what to do.
Thanks
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