- Joined
- 10 Dec 06
- Moves
- 8528
11 Nov 16
1 edit
This problem is getting in my head, becuase I can't reconcile the sloution I'm given, and the what I come to leads me to a non sensical equation. I'm rather embarassingly lost on this simple problem.
A unicyclist is cycling up a hill angled 15 degrees with respect to the horizontal. The center of mass of the cyclist is directly over the axle of the wheel and the cyclist/unicycle system have a combined mass of 100 kg. The radius of the wheel is 0.5 m and the coefficient of static friction between the wheel and the asphalt is 1. What is the magnitude of the torque the cyclist needs to exert on the pedals in order to cycle up the hill at a constant speed?
Looking forward to seeing and understanding your solution.
11 Nov 16
Originally posted by joe shmoUnless I am badly mistaken, the torque is dependant on the constant speed, so if that is not specified it cannot be solved or alternatively you can come up with an equation relating the two.
This problem is getting in my head, becuase I can't reconcile the sloution I'm given, and the what I come to leads me to a non sensical equation. I'm rather embarassingly lost on this simple problem.
A unicyclist is cycling up a hill angled 15 degrees with respect to the horizontal. The center of mass of the cyclist is directly over the axle of the whee ...[text shortened]... e up the hill at a constant speed?
Looking forward to seeing and understanding your solution.
I would say, start by working out how much energy is required to raise the cyclist over a given distance, then use that and the wheel dimensions to work out torque.
11 Nov 16
1 edit
Originally posted by twhiteheadThis looks like it works. Call the velocity uphill 1 meter/sec. P=mgv *sin A
Unless I am badly mistaken, the torque is dependant on the constant speed, so if that is not specified it cannot be solved or alternatively you can come up with an equation relating the two.
I would say, start by working out how much energy is required to raise the cyclist over a given distance, then use that and the wheel dimensions to work out torque.
A= angle uphill. So 100 kg* 9.8*1*0.259= 253 Joule/sec to go up the slope at 1 meter/sec of 15 degree slope, ignoring friction and such.
- Joined
- 10 Dec 06
- Moves
- 8528
12 Nov 16
Originally posted by twhiteheadYou may be right, that is probably why newtons laws aren't making sense ( or I'm applying them poorly)
Unless I am badly mistaken, the torque is dependant on the constant speed, so if that is not specified it cannot be solved or alternatively you can come up with an equation relating the two.
I would say, start by working out how much energy is required to raise the cyclist over a given distance, then use that and the wheel dimensions to work out torque.
Here is what I keep coming to Applying Newtons Laws to The Free body:
↑+y →+x is the standard coordinate system where the (slope) tanΦ = y/x
Now, y' and x' are normal and tangent to the slope respectively.
ΣF_y' = N - W*cosΦ = m*a_y'
N = Normal Force
W = Wieght of the rider cycle system
Φ = angle of incline
m = Mass of the rider cycle system
a_y' = acceleration of ryrder cycle system
ΣF_y' = N - W*cosΦ = m*a_y' = 0 ( a_y'=0)
N = W*cosΦ
Next:
ΣF_x' = F_fr - W*sinΦ = m*a_x' = 0 (again, a_x' = 0; velocity of body is constant)
Now, We know F_fr = µ*N = µ*W*cosΦ
Substitution into ΣF_x':
µ*W*cosΦ - W*sinΦ = 0
Leading to:
cosΦ = sinΦ which is obviously false.
So anyhow thats what I'm coming up with.
- Joined
- 10 Dec 06
- Moves
- 8528
12 Nov 16
Originally posted by twhiteheadYou know what, I'm going to have to say you are mistaken; The torque on the drive shaft is NOT dependent on the speed.
Unless I am badly mistaken, the torque is dependant on the constant speed, so if that is not specified it cannot be solved or alternatively you can come up with an equation relating the two.
I would say, start by working out how much energy is required to raise the cyclist over a given distance, then use that and the wheel dimensions to work out torque.
The torque is just the means of application of a constant force at the wheel/slope surface, that balances the component of system weight in the -x' direction . The velocity of the Unicycle/Rider system is not changing, consequently the net force acting is zero. So whatever speed I enter the slope at, I would maintain in this scenario, provided "I" ( the rider) can provide the POWER needed.
There are no velocity dependent forces in the problem.
12 Nov 16
Originally posted by joe shmoAnd you provide this POWER by means of a force, surely? And the power is proportional to the force? And the power is proportional to the slope?
So whatever speed I enter the slope at, I would maintain in this scenario, provided "I" ( the rider) can provide the POWER needed.
- Joined
- 10 Dec 06
- Moves
- 8528
12 Nov 16
Originally posted by twhiteheadYes, that is what I am saying. The body is not being accelerated, the net force acting on the body is zero.
So are you saying that if an object is being raised vertically upwards at a constant velocity, the net force acting is zero?
- Joined
- 10 Dec 06
- Moves
- 8528
13 Nov 16
3 edits
Originally posted by twhiteheadThe Force ( in this case the Torque) to climb the grade is fixed by the grade and weight of the rider /unicycle system in this scenario. It is not dependent on velocity.
And you provide this POWER by means of a force, surely? And the power is proportional to the force? And the power is proportional to the slope?
Going Back to your lifting an object vertically:
Lets say the object is "Static" (no motion) hanging by a rope. The object has a weight "W" equal to 50 N
By Newtons Second Law, the tension in the rope "T" is 50 N; The power to hold the object is 0 , because the velocity of the body is zero. P = F•v = T•0 = 0
Now attach that rope to a motor, the body is in motion vertically with constant velocity "v". Again, by Newtons Second Law the Tension in the rope "T" is 50 N. The Tension Force is independent of velocity ( neglecting air resistance, etc...), so long as the velocity is constant.
The Power is now >0: P=F•v = T•v > 0
So, in the very same respect the Force to move the body up the incline at a constant speed is independent of the speed; The Power however isn't. Hope that clears things up for you.
However, this still really isn't helping me...
13 Nov 16
1 edit
Originally posted by joe shmoWhere is this problem from? I don't believe the constraint that the centre of mass should be directly above the axle. The cyclist should be leaning forwards slightly so that the torque he applies to the wheel and, by the rotational equivalent of Newton's third law, the wheel applies to the cyclist are balanced. Otherwise the unicyclist would go over backwards. So I don't believe the physics inputs to this problem.
This problem is getting in my head, becuase I can't reconcile the sloution I'm given, and the what I come to leads me to a non sensical equation. I'm rather embarassingly lost on this simple problem.
A unicyclist is cycling up a hill angled 15 degrees with respect to the horizontal. The center of mass of the cyclist is directly over the axle of the whee ...[text shortened]... e up the hill at a constant speed?
Looking forward to seeing and understanding your solution.
The stuff about static friction is a Red Herring, the wheel rolls it doesn't slip. The work done on the wheel is the (constant) torque times angle turned through and that is the distance travelled divided by the radius of the wheel, W = T d/a where d is the distance travelled and a the radius of the wheel. The potential energy gained is mgh and h = l sin( Φ ), so we have:
T = mga sin( Φ )
13 Nov 16
Originally posted by joe shmoBut that is quite clearly intuitively wrong.
Now attach that rope to a motor, the body is in motion vertically with constant velocity "v". Again, by Newtons Second Law the Tension in the rope "T" is 50 N. The Tension Force is independent of velocity ( neglecting air resistance, etc...), so long as the velocity is constant.
The Power is now >0: P=[b]F•v = T•v > 0
So, in ...[text shortened]... er isn't. Hope that clears things up for you.
However, this still really isn't helping me...[/b]
You are saying that ignoring air resistance, a car going up a hill needs the same amount of force regardless of how fast it is going.
You clearly have never ridden a bike up a hill.
13 Nov 16
Originally posted by joe shmoΣF_x' = F_fr - W*sinΦ = m*a_x' = 0 (again, a_x' = 0; velocity of body is constant).
You may be right, that is probably why newtons laws aren't making sense ( or I'm applying them poorly)
Here is what I keep coming to Applying Newtons Laws to The Free body:
↑+y →+x is the standard coordinate system where the (slope) tanΦ = y/x
Now, y' and x' are normal and tangent to the slope respectively.
ΣF_y' = N - W*cosΦ = m*a_y'
N ...[text shortened]... Leading to:
cosΦ = sinΦ which is obviously false.
So anyhow thats what I'm coming up with.
This line is incorrect. You are trying to balance the static frictional force with the component of weight along the slope. Consider a block on a level plain, with coefficient of friction 1. The static friction is just mg. Since there are no other forces acting this frictional force should accelerate the object. Clearly this is wrong. The static friction is equal to the applied force until the applied force exceeds the maximum static friction (in this case mg), when the block will start to move. If there is no applied force then there will be no frictional force in the static case. We need to use dynamic friction then which in general will depend on speed, but for sake of simplicity assume it's equal to the maximum static friction. Then we have ma = F - mg and if there is no applied force and the block is moving then it will decelerate linearly (v = u - gt).
For your problem the relevant frictional force would be rolling friction, this is generated because the wheel has to be pulled away from a surface that it is attracted to as the wheel rolls off it. Rolling friction is not equal to static friction. I'll refer you to the Wikipedia pages entitled "friction" and "rolling resistance" for a discussion.
If you want to balance forces we have that the resistive force R is equal to cN/a, where c is the coefficient of rolling resistance (typically small of the order of 0.001), a the radius of the wheel and N the normal force. So the torque the rider has to apply to the wheel is:
T = R a = (cN/a)a = cN
Now, N is mg cos( Φ ), which gives us:
T = cmg cos( Φ )
Differing from my earlier answer, this is because we haven't taken into account the work that needs to be done to raise the weight of the bicycle up the hill. I calculated that above so the full thing is:
T = mga sin( Φ ) + mgc cos( Φ )
The first term is the torque one would need in order to overcome gravity and the second the term to take into account frictional losses.
13 Nov 16
Originally posted by twhiteheadThe problem is that the wrong frictional force was being assumed, static friction is not rolling friction. The coefficient for rolling friction I've used assumes that the wheel is rotating slowly. For higher speeds I think the coefficient (defined in the way I did in my last post) would pick up a velocity dependence, also we are neglecting friction in the axle bearings.
But that is quite clearly intuitively wrong.
You are saying that ignoring air resistance, a car going up a hill needs the same amount of force regardless of how fast it is going.
You clearly have never ridden a bike up a hill.
- Joined
- 10 Dec 06
- Moves
- 8528
13 Nov 16
1 edit
Originally posted by twhiteheadI've riden bikes up and down things you wouldn't dare imagine...but that is besides the point. This result isn't intuition someone will gain from real life experiences. This is a Physics problem ( although somewhat poorly written, in Deep Thoughts opinion). This problem is designed to show what the minimum torque needed is in the ideal case neglecting ALL forms of mechanical friction. In reality there will be losses all over the system ( some fixed, others varaible over a wide range of parameters) that will require more force at higher speeds.
But that is quite clearly intuitively wrong.
You are saying that ignoring air resistance, a car going up a hill needs the same amount of force regardless of how fast it is going.
You clearly have never ridden a bike up a hill.
However, the answer to your inquiry:
"You are saying that ignoring air resistance, a car going up a hill needs the same amount of force regardless of how fast it is going."
In the idealized (absense of mechanical friction) physical model, provided it is not accelerating (you left that bit out) is exactly...Yes
- Joined
- 10 Dec 06
- Moves
- 8528
13 Nov 16
2 edits
Originally posted by DeepThought"Where is this problem from? I don't believe the constraint that the centre of mass should be directly above the axle. The cyclist should be leaning forwards slightly so that the torque he applies to the wheel and, by the rotational equivalent of Newton's third law, the wheel applies to the cyclist are balanced. Otherwise the unicyclist would go over backwards. So I don't believe the physics inputs to this problem."
Where is this problem from? I don't believe the constraint that the centre of mass should be directly above the axle. The cyclist should be leaning forwards slightly so that the torque he applies to the wheel and, by the rotational equivalent of Newton's third law, the wheel applies to the cyclist are balanced. Otherwise the unicyclist would go over b ...[text shortened]... e wheel. The potential energy gained is mgh and h = l sin( Φ ), so we have:
T = mga sin( Φ )
The problem was from: https://brilliant.org/
I now see and agree with you. There is an unbalanced torque on the body if this center of mass constraint is imposed. The center of mass has to be offset to the +x direction for this scenario to be possible.
"The stuff about static friction is a Red Herring, the wheel rolls it doesn't slip. The work done on the wheel is the (constant) torque times angle turned through and that is the distance travelled divided by the radius of the wheel, W = T d/a where d is the distance travelled and a the radius of the wheel. The potential energy gained is mgh and h = l sin( Φ ), so we have:"
Ok, so neglecting rolling resistance: The forces on the wheel in the FBD are The normal force ( +y' ) and the Reaction Force from the driving force ( +x' ) and the systems weight ( -y ). The solution on the site literally was saying that static friction force was driving the unicyclist up the hill,... I took the bait and tried to reconcile it (arriving at a contradiction).
Thank you for your help on this!