Mathmatical question (A Level)
Discussion
I'm studying Mechanics / Motor sport and the degree which due to the motor sport means A Level equivalent math.
We've been set this question (I'll start you with one for now) which we haven't actually been taught.
We need to research it ourselves but I don't even know what to Google.
The question is as follows
We've been set this question (I'll start you with one for now) which we haven't actually been taught.
We need to research it ourselves but I don't even know what to Google.
The question is as follows
Math Question said:
The road speed of a vehicle increases the speed uniformly from 500 rev/min to 1100 rev/min in 40 seconds. The overall diameter of the wheel and tyre is 0.7m.
(Rad = Radians)
i) Calculate the angular acceleration of the road wheel in rad/s².
ii) Calculate the linear acceleration of a point on the tyre tread.
So if anyone knows the equation / can help at all please do!(Rad = Radians)
i) Calculate the angular acceleration of the road wheel in rad/s².
ii) Calculate the linear acceleration of a point on the tyre tread.
The first part is pretty easy, the second part might be straightforward, or more complex.
I'd imagine that the second part is seeking an expression that defines the locus of the point on the tyre tread, a combination of the linear, and angular acceleration of the wheel.
If it's a degree question, answer the first two questions the easy way, and then answer the implied third question that does not actually appear.
I'd imagine that the second part is seeking an expression that defines the locus of the point on the tyre tread, a combination of the linear, and angular acceleration of the wheel.
If it's a degree question, answer the first two questions the easy way, and then answer the implied third question that does not actually appear.
Looks just about as confusing as I thought then...
Think this will take a bit longer than the usual 30 minutes of Googling / Badgering someone else and bull s
tting it into actual text.
I'll leave this one until tomorrow and see if I've got anything easier to do in the mean time
Think this will take a bit longer than the usual 30 minutes of Googling / Badgering someone else and bull s
tting it into actual text.I'll leave this one until tomorrow and see if I've got anything easier to do in the mean time
Edited by MGB Boy on Friday 9th April 15:18
MGB Boy said:
I'm studying Mechanics / Motor sport and the degree which due to the motor sport means A Level equivalent math.
We've been set this question (I'll start you with one for now) which we haven't actually been taught.
We need to research it ourselves but I don't even know what to Google.
The question is as follows
Easy. The answer's yes.We've been set this question (I'll start you with one for now) which we haven't actually been taught.
We need to research it ourselves but I don't even know what to Google.
The question is as follows
Math Question said:
The road speed of a vehicle increases the speed uniformly from 500 rev/min to 1100 rev/min in 40 seconds. The overall diameter of the wheel and tyre is 0.7m.
(Rad = Radians)
i) Calculate the angular acceleration of the road wheel in rad/s².
ii) Calculate the linear acceleration of a point on the tyre tread.
So if anyone knows the equation / can help at all please do!(Rad = Radians)
i) Calculate the angular acceleration of the road wheel in rad/s².
ii) Calculate the linear acceleration of a point on the tyre tread.
What's the angular veolicty at the start (in rads per sec) and the angular velocity at the end ?
The difference is the increase in angular velocity over that time period. Divide the former by the latter to get the angular acceleration.
Actually the question is written incorrectly since the road speed of the vehicle is not measured in revs per minute. The engine speed might be.
To go from revs per minute to rad per sec, divide revs by 180 and multiply by pi remembering to converting minutes to seconds as well.
The second part is ambiguous, linear acceleration relative to what ? relative to the centre of teh wheel the acceleration is zero. I suspect it means relative to a point on the road or other such non rotating point.
have a look at your notes on simple harminc motion and you'll see the forumla for velocity relative to a fixed point, and it's a since wave. Since accelaration is the first derivative of velocity it'll look like a cosine wave.
The difference is the increase in angular velocity over that time period. Divide the former by the latter to get the angular acceleration.
Actually the question is written incorrectly since the road speed of the vehicle is not measured in revs per minute. The engine speed might be.
To go from revs per minute to rad per sec, divide revs by 180 and multiply by pi remembering to converting minutes to seconds as well.
The second part is ambiguous, linear acceleration relative to what ? relative to the centre of teh wheel the acceleration is zero. I suspect it means relative to a point on the road or other such non rotating point.
have a look at your notes on simple harminc motion and you'll see the forumla for velocity relative to a fixed point, and it's a since wave. Since accelaration is the first derivative of velocity it'll look like a cosine wave.
Edited by 911newbie on Friday 9th April 16:36
911newbie said:
To go from revs per minute to rad per sec, divide revs by 180 and multiply by pi remembering to converting minutes to seconds as well.
Its been a while but don't you just times Revs / min by 2 x pi to get rads / min. Then divide the answer by 60 to get rads / sec ?Convert both to Rads / sec subtract lower from higher divide by time and that should answer part i)
For part ii) you probably have to work out the circumference of the wheel (pi x diameter) and the linear distance covered by the wheel at the start and finish speeds. I have not thought about part ii) properly, or part i) for that matter but think I am barking up the right tree.
Ben
esuuv said:
Friday afternoon and my brains fried so no doubt someone will be along in a second to prove me wrong but I make it......
1.56 rad s -2
0.546 m s -2
(How do you do write superscript in here?)
I get the same/similar1.56 rad s -2
0.546 m s -2
(How do you do write superscript in here?)
Angular acceleration is simply PI/2 = roughly 1.57 rad/s^2
Linear Acceleration is 7PI/40 = roughly 0.55 rad/s^2
aa = 2PI * (1100-500) / (60*40)
aa = 2PI * 600/(60*40)
aa = 2PI * 10/40
aa = 2PI *1/4
aa = PI/2
la = 0.7PI * (1100-500) / (60*40)
la = 0.7PI * 600/(60*40)
la = 0.7PI * 10/40
la = 7PI/40
Given how dumbed-down A-levels are these days 
the second part is most likely to be a simple case of converting the angular velocity and radius into a linear speed component and measuring the acceleration of that. I can't see it being a locus or anything involving more than just the most basic of formulae.
In which case, assuming that the angular measurements given in the question are those of the wheel itself (this is ambiguous):
i) The angular velocity changes by 600rpm in 40 secs. This is equivalent to 10 revs per second in 40 seconds, or 1/4 rev per second squared. One rev is 2*pi radians, so the answer is pi/2 rads per second squared (or approximately 1.5708).
ii) The diameter is given as 0.7m; the circumference is this multiplied by pi. One revolution is therefore equivalent to a distance of (0.7 pi). We've already established that the acceleration is 1/4 of a revolution per second squared, hence it is (0.7 pi)/4 metres per second squared, or 0.55 approx.
esuuv had correct answers (but with a rough value for pi?)
the second part is most likely to be a simple case of converting the angular velocity and radius into a linear speed component and measuring the acceleration of that. I can't see it being a locus or anything involving more than just the most basic of formulae.In which case, assuming that the angular measurements given in the question are those of the wheel itself (this is ambiguous):
i) The angular velocity changes by 600rpm in 40 secs. This is equivalent to 10 revs per second in 40 seconds, or 1/4 rev per second squared. One rev is 2*pi radians, so the answer is pi/2 rads per second squared (or approximately 1.5708).
ii) The diameter is given as 0.7m; the circumference is this multiplied by pi. One revolution is therefore equivalent to a distance of (0.7 pi). We've already established that the acceleration is 1/4 of a revolution per second squared, hence it is (0.7 pi)/4 metres per second squared, or 0.55 approx.
esuuv had correct answers (but with a rough value for pi?)
Edited by moosepig on Friday 9th April 16:55
the-gofer said:
esuuv said:
Friday afternoon and my brains fried so no doubt someone will be along in a second to prove me wrong but I make it......
1.56 rad s -2
0.546 m s -2
(How do you do write superscript in here?)
I get the same/similar1.56 rad s -2
0.546 m s -2
(How do you do write superscript in here?)
Angular acceleration is simply PI/2 = roughly 1.57 rad/s^2
Linear Acceleration is 7PI/40 = roughly 0.55 rad/s^2
aa = 2PI * (1100-500) / (60*40)
aa = 2PI * 600/(60*40)
aa = 2PI * 10/40
aa = 2PI *1/4
aa = PI/2
la = 0.7PI * (1100-500) / (60*40)
la = 0.7PI * 600/(60*40)
la = 0.7PI * 10/40
la = 7PI/40
Edited to add: I was always taught at Maths A-level that the question writers pick numbers that usually end up in neat fractions - if it doesn't end in neat fractions you've interpreted the question wrong (or differently
) or you've made a mistake somewhere. The fact the the Angular Velocity comes out as PI/2 and the Linear Acceleration comes out as a neat 7/40 fraction of PI suggests that this is indeed the answer they are looking for and that they therefore mean the acceleration is relative to a tangent from the edge of the circle - i.e. how fast would the vehicle accelerate in real terms. Which, if you think about it is the obvious question to ask.Edited by the-gofer on Friday 9th April 16:58
Feel free to correct me - it's too late on a Friday afternoon. I make it:
i) a = (v-u)/t where a is angular acceleration, v is ultimate angular velocity, u is initial angular velocity and t is time.
u = 500rev/min = 500/60 rev/s = 8.3rev/s.
there are 2pi radians in one revolution so u = 8.3*2pi rad/s = 52.4rad/s
v = 1100rev/min = 1100/60 rev/s = 18.3rev/s = 18.3*2pi rad/s = 115.2rad/s
SO, a = (115.2-52.4)/40 = 62.8/40 = 1.57 rad/s^2
ii) a full rotation has an angle of 2pi radians and circumference of 2pi r where r is the radius of the wheel.
so A = a*r where A is linear acceleration and r is the radius of the wheel
A=1.57*0.35m = 0.55 m/s
i) a = (v-u)/t where a is angular acceleration, v is ultimate angular velocity, u is initial angular velocity and t is time.
u = 500rev/min = 500/60 rev/s = 8.3rev/s.
there are 2pi radians in one revolution so u = 8.3*2pi rad/s = 52.4rad/s
v = 1100rev/min = 1100/60 rev/s = 18.3rev/s = 18.3*2pi rad/s = 115.2rad/s
SO, a = (115.2-52.4)/40 = 62.8/40 = 1.57 rad/s^2
ii) a full rotation has an angle of 2pi radians and circumference of 2pi r where r is the radius of the wheel.
so A = a*r where A is linear acceleration and r is the radius of the wheel
A=1.57*0.35m = 0.55 m/s
sneijder said:
If it makes you feel any better, one of my mates started this very course. He thought he'd be on the Ferrari F1 pit crew in a couple of years.
Then the Maths bomb dropped.
If you can't cope with it, maybe a bit of extra tuition is needed ?
Oh, and the answer is indeed 7.
I also know someone who started the degree then dropped out.Then the Maths bomb dropped.
If you can't cope with it, maybe a bit of extra tuition is needed ?
Oh, and the answer is indeed 7.
But it's a bit harsh saying that he can't cope with it - a degree it meant to be difficult and make you learn, if he knew the answer straight away without being taught the theory behind it then he'd be a genius.
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