best spin bike value image
Q. What is it?
What is it used for?
What does it explain?
And how do you find it mathematically?
What is it used for?
What does it explain?
And how do you find it mathematically?
Answer
The moment of inertia of a body is â m(i) . r(i)^2
Which means that you take all the individual masses that make up the body , multiply each one by the square of the distance it is from the axis of rotation and add them all together.
For each body there can be any number of axes of rotation and so there can be any number of different Moments of Inertia for the same body.
We concentrate on simple shapes with axes of rotation in obvious places.
For example - a bike wheel with all the mass in the rim ( zero weight spokes ! ) , and every bit of that mass the same distance R from the centre of the circle . So the MI of the wheel is M . R^2
Next step = A solid disk ( like a CD ) , spinning round its centre. This time some of the mass is closer to the
centre, and contributes less to the MI .
The total MI for a solid disc can be worked out ( it needs calculus ) and it comes to 1/2 M R^2
Another way to write this would be Mr^2 where r is the radius of gyration - the single value for the average distance from the rotation axis that will give you the correct moment of inertia
In this case r = R/â2
All moments of inertia can be represented by the formula MI = M . r^2 where the r ( the radius of gyration ) is the representative distance that gives you the correct answer.
That is all it is.
A solid ball spinning about a central axis, has a moment of inertia of 2/5 . M . R^2
so its radius of gyration will be 0.4 . M . R^2 = M . r^2
r = R . â 0.4 = 0.6325 R
You don't usually work out the radius of gyration.... you measure it practically by seeing what torque is needed to spin an object, work out the moment of inertia ( by experiment ) and then calculate what the gyration radius must be.
Unless you are doing engineering, I would just ignore it.
The moment of inertia of a body is â m(i) . r(i)^2
Which means that you take all the individual masses that make up the body , multiply each one by the square of the distance it is from the axis of rotation and add them all together.
For each body there can be any number of axes of rotation and so there can be any number of different Moments of Inertia for the same body.
We concentrate on simple shapes with axes of rotation in obvious places.
For example - a bike wheel with all the mass in the rim ( zero weight spokes ! ) , and every bit of that mass the same distance R from the centre of the circle . So the MI of the wheel is M . R^2
Next step = A solid disk ( like a CD ) , spinning round its centre. This time some of the mass is closer to the
centre, and contributes less to the MI .
The total MI for a solid disc can be worked out ( it needs calculus ) and it comes to 1/2 M R^2
Another way to write this would be Mr^2 where r is the radius of gyration - the single value for the average distance from the rotation axis that will give you the correct moment of inertia
In this case r = R/â2
All moments of inertia can be represented by the formula MI = M . r^2 where the r ( the radius of gyration ) is the representative distance that gives you the correct answer.
That is all it is.
A solid ball spinning about a central axis, has a moment of inertia of 2/5 . M . R^2
so its radius of gyration will be 0.4 . M . R^2 = M . r^2
r = R . â 0.4 = 0.6325 R
You don't usually work out the radius of gyration.... you measure it practically by seeing what torque is needed to spin an object, work out the moment of inertia ( by experiment ) and then calculate what the gyration radius must be.
Unless you are doing engineering, I would just ignore it.
Physics 11 Questions Redgarding Force of Friction PLease Help?
Wesley
A Guy and his bike have total mass of 67.0 kg. The cyclist is travelling at constant speed along horizontal highway. As he peddles, the forward force exerted on him is 154 N
What does it mean when they say forward force exerted on him? is that going forwards or backwards?
And how can I find out friction force and the coefficent?
Answer
You are asking a very difficult (maybe impossible) question. Here is some background information, which may (or may not!) help.
Assuming the cyclist is also going in a straight line, there will be 3 horizontal forces on him:
a) 2 forces in the opposite direction to his velocity; these are air resistance and also what is called 'rolling resistance' - their total will equal 154N. (If you are not familiar with rolling resistance, just imagine the wheel-bearings need greasing and/or the tyres are flat - these factors would increase rolling resistance a lot!)
b) a force (call it 'thrust') in the direction of motion; this will also be 154N; this is the forwards frictional force of the ground on the driving-wheel. (This is a difficult concept; the thrust is the 'grip' between the wheel and road and is stopping the rotating wheel from sliding against the road as it rotates - as a consequence it thrusts the wheel forwards.)
a) and b) have opposite directions; as the cyclist travels at constant velocity we know the resultant of a) and b) is zero, hence they must both have a magnitude of 154N
Which of these forces is called the 'forwards force' not 100% clear, but I would opt for b), the thrust.
The confusion here is that there are 3 frictional forces; air resistance and rolling resistance are frictional and act backwards. But the thrust is also a frictional force and acts forwards.
There are different types of coefficient of friction:
- the coefficient of static friction applies to surfaces just about to start slipping (the case of limiting friction) and is not applicable here unless wheel-spin is about to start;
- coefficient of kinetic friction applies to surfaces which are already sliding against each other; again this is not applicable here;
- coefficient of rolling friction could be applied but should not include air resistance, , so there is no way to work it out here.
If you actually need a value for the coefficient of friction (which you should NOT have been asked for as it makes no sense in this situation) you would have to invent an 'effective coefficient of frictionâ which included the effects of air resistance; this would be calculated as follows:
(total fictional force opposing motion) / (normal force)
= 154 /(67.0 x 9.81)
= 0.234
Hope that helps a bit.
You are asking a very difficult (maybe impossible) question. Here is some background information, which may (or may not!) help.
Assuming the cyclist is also going in a straight line, there will be 3 horizontal forces on him:
a) 2 forces in the opposite direction to his velocity; these are air resistance and also what is called 'rolling resistance' - their total will equal 154N. (If you are not familiar with rolling resistance, just imagine the wheel-bearings need greasing and/or the tyres are flat - these factors would increase rolling resistance a lot!)
b) a force (call it 'thrust') in the direction of motion; this will also be 154N; this is the forwards frictional force of the ground on the driving-wheel. (This is a difficult concept; the thrust is the 'grip' between the wheel and road and is stopping the rotating wheel from sliding against the road as it rotates - as a consequence it thrusts the wheel forwards.)
a) and b) have opposite directions; as the cyclist travels at constant velocity we know the resultant of a) and b) is zero, hence they must both have a magnitude of 154N
Which of these forces is called the 'forwards force' not 100% clear, but I would opt for b), the thrust.
The confusion here is that there are 3 frictional forces; air resistance and rolling resistance are frictional and act backwards. But the thrust is also a frictional force and acts forwards.
There are different types of coefficient of friction:
- the coefficient of static friction applies to surfaces just about to start slipping (the case of limiting friction) and is not applicable here unless wheel-spin is about to start;
- coefficient of kinetic friction applies to surfaces which are already sliding against each other; again this is not applicable here;
- coefficient of rolling friction could be applied but should not include air resistance, , so there is no way to work it out here.
If you actually need a value for the coefficient of friction (which you should NOT have been asked for as it makes no sense in this situation) you would have to invent an 'effective coefficient of frictionâ which included the effects of air resistance; this would be calculated as follows:
(total fictional force opposing motion) / (normal force)
= 154 /(67.0 x 9.81)
= 0.234
Hope that helps a bit.
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