How do you find the speed of a car without skidding?Circular MotionFind coefficient of static friction if...
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How do you find the speed of a car without skidding?
Circular MotionFind coefficient of static friction if given initial velocity and distance?Angle that car is at after angular accelerationUniform Circular Motion with Banked Road and CarRolling a ball into a cone; what should the forces overall be?The radius of the track.How to calculate coefficient of static friction given certain dataconfusion on force = mass x accelerationDetermining kinetic friction coefficientBouncing ball with friction.
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Q.)A car rounds a turn of radius 120 m on a flat road. If the coefficient of friction between the tires and
the road is 0.50, what is the maximum speed of the car without skidding?
I really don't get this problem. I get that centripetal force= mass*(velocity^2/radius) and frictional force = coefficient of friction*Normal force. If I put 120 m as radius for centripetal force I get just v^2 multiplied by a number which im trying to find(thats fine there), and if I put 0.50 as a sub for the coefficient I just have N or mg mutiplied by 0.50.
If I put it in newton's 2nd law being F(centripetal)-F(friction)=ma,
it wouldn't make sense because the acceleration and velocity is still undefined. Can anyone help me tackle this problem? Or am I just way off track?
physics
asked Apr 14 '17 at 21:51
NemesisNemesis
1314
$endgroup$
add a comment |
$begingroup$
Q.)A car rounds a turn of radius 120 m on a flat road. If the coefficient of friction between the tires and
the road is 0.50, what is the maximum speed of the car without skidding?
I really don't get this problem. I get that centripetal force= mass*(velocity^2/radius) and frictional force = coefficient of friction*Normal force. If I put 120 m as radius for centripetal force I get just v^2 multiplied by a number which im trying to find(thats fine there), and if I put 0.50 as a sub for the coefficient I just have N or mg mutiplied by 0.50.
If I put it in newton's 2nd law being F(centripetal)-F(friction)=ma,
it wouldn't make sense because the acceleration and velocity is still undefined. Can anyone help me tackle this problem? Or am I just way off track?
physics
asked Apr 14 '17 at 21:51
NemesisNemesis
1314
$endgroup$
$begingroup$
Hi and welcome to the site. There is a typesetting called MathJax which uses LaTeX which is good to learn on the site. Other than that it is very good that you show any own attempts at solving problems, especially if they are homework of some kind.
$endgroup$
– mathreadler
Apr 14 '17 at 22:15
$begingroup$
Was the question originally stated in English? It's ended up with an MSE title that probably doesn't quite mean what was intended $ddot{smile}$: I think I can find the answer while seated safely in my armchair! I leave how to typeset $ddot{smile}$ as a $LaTeX$ exercise.
$endgroup$
– Rob Arthan
Apr 14 '17 at 22:15
add a comment |
$begingroup$
Q.)A car rounds a turn of radius 120 m on a flat road. If the coefficient of friction between the tires and
the road is 0.50, what is the maximum speed of the car without skidding?
I really don't get this problem. I get that centripetal force= mass*(velocity^2/radius) and frictional force = coefficient of friction*Normal force. If I put 120 m as radius for centripetal force I get just v^2 multiplied by a number which im trying to find(thats fine there), and if I put 0.50 as a sub for the coefficient I just have N or mg mutiplied by 0.50.
If I put it in newton's 2nd law being F(centripetal)-F(friction)=ma,
it wouldn't make sense because the acceleration and velocity is still undefined. Can anyone help me tackle this problem? Or am I just way off track?
physics
asked Apr 14 '17 at 21:51
NemesisNemesis
1314
$endgroup$
Q.)A car rounds a turn of radius 120 m on a flat road. If the coefficient of friction between the tires and
the road is 0.50, what is the maximum speed of the car without skidding?
I really don't get this problem. I get that centripetal force= mass*(velocity^2/radius) and frictional force = coefficient of friction*Normal force. If I put 120 m as radius for centripetal force I get just v^2 multiplied by a number which im trying to find(thats fine there), and if I put 0.50 as a sub for the coefficient I just have N or mg mutiplied by 0.50.
If I put it in newton's 2nd law being F(centripetal)-F(friction)=ma,
it wouldn't make sense because the acceleration and velocity is still undefined. Can anyone help me tackle this problem? Or am I just way off track?
physics
physics
asked Apr 14 '17 at 21:51
NemesisNemesis
1314
asked Apr 14 '17 at 21:51
NemesisNemesis
1314
asked Apr 14 '17 at 21:51
NemesisNemesis
1314
asked Apr 14 '17 at 21:51
NemesisNemesis
1314
asked Apr 14 '17 at 21:51
NemesisNemesis
1314
1314
$begingroup$
Hi and welcome to the site. There is a typesetting called MathJax which uses LaTeX which is good to learn on the site. Other than that it is very good that you show any own attempts at solving problems, especially if they are homework of some kind.
$endgroup$
– mathreadler
Apr 14 '17 at 22:15
$begingroup$
Was the question originally stated in English? It's ended up with an MSE title that probably doesn't quite mean what was intended $ddot{smile}$: I think I can find the answer while seated safely in my armchair! I leave how to typeset $ddot{smile}$ as a $LaTeX$ exercise.
$endgroup$
– Rob Arthan
Apr 14 '17 at 22:15
add a comment |
$begingroup$
Hi and welcome to the site. There is a typesetting called MathJax which uses LaTeX which is good to learn on the site. Other than that it is very good that you show any own attempts at solving problems, especially if they are homework of some kind.
$endgroup$
– mathreadler
Apr 14 '17 at 22:15
$begingroup$
Was the question originally stated in English? It's ended up with an MSE title that probably doesn't quite mean what was intended $ddot{smile}$: I think I can find the answer while seated safely in my armchair! I leave how to typeset $ddot{smile}$ as a $LaTeX$ exercise.
$endgroup$
– Rob Arthan
Apr 14 '17 at 22:15
$begingroup$
Hi and welcome to the site. There is a typesetting called MathJax which uses LaTeX which is good to learn on the site. Other than that it is very good that you show any own attempts at solving problems, especially if they are homework of some kind.
$endgroup$
– mathreadler
Apr 14 '17 at 22:15
$begingroup$
Hi and welcome to the site. There is a typesetting called MathJax which uses LaTeX which is good to learn on the site. Other than that it is very good that you show any own attempts at solving problems, especially if they are homework of some kind.
$endgroup$
– mathreadler
Apr 14 '17 at 22:15
$begingroup$
Was the question originally stated in English? It's ended up with an MSE title that probably doesn't quite mean what was intended $ddot{smile}$: I think I can find the answer while seated safely in my armchair! I leave how to typeset $ddot{smile}$ as a $LaTeX$ exercise.
$endgroup$
– Rob Arthan
Apr 14 '17 at 22:15
$begingroup$
Was the question originally stated in English? It's ended up with an MSE title that probably doesn't quite mean what was intended $ddot{smile}$: I think I can find the answer while seated safely in my armchair! I leave how to typeset $ddot{smile}$ as a $LaTeX$ exercise.
$endgroup$
– Rob Arthan
Apr 14 '17 at 22:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The centripetal motion is supplied by the force of friction $f = mu N = mu mg$. Setting equal
$$ frac{mv^2}{r} = mu mg to v = sqrt{mu g r} $$
answered Apr 14 '17 at 21:56
GregoryGregory
1,346412
$endgroup$
$begingroup$
Oh! It was like that. Thanks you!
$endgroup$
– Nemesis
Apr 14 '17 at 22:00
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
The centripetal motion is supplied by the force of friction $f = mu N = mu mg$. Setting equal
$$ frac{mv^2}{r} = mu mg to v = sqrt{mu g r} $$
answered Apr 14 '17 at 21:56
GregoryGregory
1,346412
$endgroup$
$begingroup$
Oh! It was like that. Thanks you!
$endgroup$
– Nemesis
Apr 14 '17 at 22:00
add a comment |
$begingroup$
The centripetal motion is supplied by the force of friction $f = mu N = mu mg$. Setting equal
$$ frac{mv^2}{r} = mu mg to v = sqrt{mu g r} $$
answered Apr 14 '17 at 21:56
GregoryGregory
1,346412
$endgroup$
$begingroup$
Oh! It was like that. Thanks you!
$endgroup$
– Nemesis
Apr 14 '17 at 22:00
add a comment |
$begingroup$
The centripetal motion is supplied by the force of friction $f = mu N = mu mg$. Setting equal
$$ frac{mv^2}{r} = mu mg to v = sqrt{mu g r} $$
answered Apr 14 '17 at 21:56
GregoryGregory
1,346412
$endgroup$
The centripetal motion is supplied by the force of friction $f = mu N = mu mg$. Setting equal
$$ frac{mv^2}{r} = mu mg to v = sqrt{mu g r} $$
answered Apr 14 '17 at 21:56
GregoryGregory
1,346412
answered Apr 14 '17 at 21:56
GregoryGregory
1,346412
answered Apr 14 '17 at 21:56
GregoryGregory
1,346412
answered Apr 14 '17 at 21:56
GregoryGregory
1,346412
1,346412
$begingroup$
Oh! It was like that. Thanks you!
$endgroup$
– Nemesis
Apr 14 '17 at 22:00
add a comment |
$begingroup$
Oh! It was like that. Thanks you!
$endgroup$
– Nemesis
Apr 14 '17 at 22:00
$begingroup$
Oh! It was like that. Thanks you!
$endgroup$
– Nemesis
Apr 14 '17 at 22:00
$begingroup$
Oh! It was like that. Thanks you!
$endgroup$
– Nemesis
Apr 14 '17 at 22:00
add a comment |
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$begingroup$
Hi and welcome to the site. There is a typesetting called MathJax which uses LaTeX which is good to learn on the site. Other than that it is very good that you show any own attempts at solving problems, especially if they are homework of some kind.
$endgroup$
– mathreadler
Apr 14 '17 at 22:15
$begingroup$
Was the question originally stated in English? It's ended up with an MSE title that probably doesn't quite mean what was intended $ddot{smile}$: I think I can find the answer while seated safely in my armchair! I leave how to typeset $ddot{smile}$ as a $LaTeX$ exercise.
$endgroup$
– Rob Arthan
Apr 14 '17 at 22:15