The 1100-kg mass of a car includes four tires, each of mass(including wheels) 35
ID: 1671433 • Letter: T
Question
The 1100-kg mass of a car includes four tires, each of mass(including wheels) 35kg and diameter 0.80m. Assume each tire andwheel combination acts as a solid cylinder.Determine
(a) the total kinetic energy of the car when traveling 95km/h
(b)the fraction of the kinetic energy in the tires and wheels
(c) If the car is initially at rest and is then pulled by a towtruck with a force of 1500N, what is the acceleration of the car?(Ignore frictional losses)
(d) What % error would you make in part (c) if you ignored therotational inertia of the tires and wheels?
Explanation / Answer
Question Details: The 1100-kg mass of a car includes four tires, each of mass(including wheels) 35kg and diameter 0.80m. Assume each tire andwheel combination acts as a solid cylinder.Determine
(a) the total kinetic energy of the car when traveling 95km/h
[1] Total kinetic energy = translation + rotation =1/2mv^2+1/2I^2
The velocity converts to v=26.4m/s
I for a solid cylinder= 1/2Mr^2, which for ourcase is 1/2(35kg)(.40m)^2=I=2.8kgm^2
To find for a tire, we need to use(=v/r), =(26.4m/s)/(.40m)=65.97 or just=66rad/s To find the total energy, use eq[1] and plug in all the valueswe just found, since there are four tires, multiply I^2 by4 Total energy=1/2(1100kg)(26.4m/s)^2 +2(2.8kgm^2)(66rad/s)^2=407721Joules or just Totalenergy=410,000J
(b)the fraction of the kinetic energy in the tires and wheels KE=1/2mv^2 There are 4 tires whose combined mass is 4*(35kg)=140kg KEtires=.5(140kg)(26.4m/s)^2=48787J The fraction of the KE in tires = 48787J/407721J=.1197 or 12%or 3/25
(c) If the car is initially at rest and is then pulled by a towtruck with a force of 1500N, what is the acceleration of the car?(Ignore frictional losses) The car goes from having a total kinetic energy of 0J tohaving a TKE of 407721J Which means that the tow truck would have to do that amount ofwork to give the car that energy Since work=forcexdistance, we can figure out the distance byusing the energy and force we've been given 407721J=1500Nxdistance means that distance=271.8m Now that we know speed(26.4m/s) and distance(271.8m) we canfind acceleration by saying that final velocity, v=2(distance)/(time) , plugging in ourconstants, we get 26.4m/s=2(271.8m)/time and therefore, since time=20.6s, theacceleration lasted for 20.6s Then we use acceleration, a=v/time, and plug in our constantsto get a=(26.4m/s)/(20.6s)=a=1.28m/s^2
(d) What % error would you make in part (c) if you ignored therotational inertia of the tires and wheels? kinetic energy from rotation(revisited), KE=I^2, wasKE=4x.5(2.8kgm^2)(66rad/s)^2=24394J or KE=24,000J Since the ratio of the rotational intertia to the Total KE is24,000/410,000=.059 or 5.9% I would use 6% errorto have a good round number to work with. Question Details: The 1100-kg mass of a car includes four tires, each of mass(including wheels) 35kg and diameter 0.80m. Assume each tire andwheel combination acts as a solid cylinder.
Determine
(a) the total kinetic energy of the car when traveling 95km/h
[1] Total kinetic energy = translation + rotation =1/2mv^2+1/2I^2
The velocity converts to v=26.4m/s
I for a solid cylinder= 1/2Mr^2, which for ourcase is 1/2(35kg)(.40m)^2=I=2.8kgm^2
To find for a tire, we need to use(=v/r), =(26.4m/s)/(.40m)=65.97 or just=66rad/s To find the total energy, use eq[1] and plug in all the valueswe just found, since there are four tires, multiply I^2 by4 Total energy=1/2(1100kg)(26.4m/s)^2 +2(2.8kgm^2)(66rad/s)^2=407721Joules or just Totalenergy=410,000J
(b)the fraction of the kinetic energy in the tires and wheels KE=1/2mv^2 There are 4 tires whose combined mass is 4*(35kg)=140kg KEtires=.5(140kg)(26.4m/s)^2=48787J The fraction of the KE in tires = 48787J/407721J=.1197 or 12%or 3/25
(c) If the car is initially at rest and is then pulled by a towtruck with a force of 1500N, what is the acceleration of the car?(Ignore frictional losses) The car goes from having a total kinetic energy of 0J tohaving a TKE of 407721J Which means that the tow truck would have to do that amount ofwork to give the car that energy Since work=forcexdistance, we can figure out the distance byusing the energy and force we've been given 407721J=1500Nxdistance means that distance=271.8m Now that we know speed(26.4m/s) and distance(271.8m) we canfind acceleration by saying that final velocity, v=2(distance)/(time) , plugging in ourconstants, we get 26.4m/s=2(271.8m)/time and therefore, since time=20.6s, theacceleration lasted for 20.6s Then we use acceleration, a=v/time, and plug in our constantsto get a=(26.4m/s)/(20.6s)=a=1.28m/s^2
(d) What % error would you make in part (c) if you ignored therotational inertia of the tires and wheels? kinetic energy from rotation(revisited), KE=I^2, wasKE=4x.5(2.8kgm^2)(66rad/s)^2=24394J or KE=24,000J Since the ratio of the rotational intertia to the Total KE is24,000/410,000=.059 or 5.9% I would use 6% errorto have a good round number to work with. Question Details: The 1100-kg mass of a car includes four tires, each of mass(including wheels) 35kg and diameter 0.80m. Assume each tire andwheel combination acts as a solid cylinder.
Determine
(a) the total kinetic energy of the car when traveling 95km/h
[1] Total kinetic energy = translation + rotation =1/2mv^2+1/2I^2
The velocity converts to v=26.4m/s
I for a solid cylinder= 1/2Mr^2, which for ourcase is 1/2(35kg)(.40m)^2=I=2.8kgm^2
To find for a tire, we need to use(=v/r), =(26.4m/s)/(.40m)=65.97 or just=66rad/s To find the total energy, use eq[1] and plug in all the valueswe just found, since there are four tires, multiply I^2 by4 Total energy=1/2(1100kg)(26.4m/s)^2 +2(2.8kgm^2)(66rad/s)^2=407721Joules or just Totalenergy=410,000J
(b)the fraction of the kinetic energy in the tires and wheels KE=1/2mv^2 There are 4 tires whose combined mass is 4*(35kg)=140kg KEtires=.5(140kg)(26.4m/s)^2=48787J The fraction of the KE in tires = 48787J/407721J=.1197 or 12%or 3/25
(c) If the car is initially at rest and is then pulled by a towtruck with a force of 1500N, what is the acceleration of the car?(Ignore frictional losses) The car goes from having a total kinetic energy of 0J tohaving a TKE of 407721J Which means that the tow truck would have to do that amount ofwork to give the car that energy Since work=forcexdistance, we can figure out the distance byusing the energy and force we've been given 407721J=1500Nxdistance means that distance=271.8m Now that we know speed(26.4m/s) and distance(271.8m) we canfind acceleration by saying that final velocity, v=2(distance)/(time) , plugging in ourconstants, we get 26.4m/s=2(271.8m)/time and therefore, since time=20.6s, theacceleration lasted for 20.6s Then we use acceleration, a=v/time, and plug in our constantsto get a=(26.4m/s)/(20.6s)=a=1.28m/s^2
(d) What % error would you make in part (c) if you ignored therotational inertia of the tires and wheels? kinetic energy from rotation(revisited), KE=I^2, wasKE=4x.5(2.8kgm^2)(66rad/s)^2=24394J or KE=24,000J Since the ratio of the rotational intertia to the Total KE is24,000/410,000=.059 or 5.9% I would use 6% errorto have a good round number to work with. The 1100-kg mass of a car includes four tires, each of mass(including wheels) 35kg and diameter 0.80m. Assume each tire andwheel combination acts as a solid cylinder.
Determine
(a) the total kinetic energy of the car when traveling 95km/h
[1] Total kinetic energy = translation + rotation =1/2mv^2+1/2I^2
The velocity converts to v=26.4m/s
I for a solid cylinder= 1/2Mr^2, which for ourcase is 1/2(35kg)(.40m)^2=I=2.8kgm^2
To find for a tire, we need to use(=v/r), =(26.4m/s)/(.40m)=65.97 or just=66rad/s To find the total energy, use eq[1] and plug in all the valueswe just found, since there are four tires, multiply I^2 by4 Total energy=1/2(1100kg)(26.4m/s)^2 +2(2.8kgm^2)(66rad/s)^2=407721Joules or just Totalenergy=410,000J
(b)the fraction of the kinetic energy in the tires and wheels KE=1/2mv^2 There are 4 tires whose combined mass is 4*(35kg)=140kg KEtires=.5(140kg)(26.4m/s)^2=48787J The fraction of the KE in tires = 48787J/407721J=.1197 or 12%or 3/25
(c) If the car is initially at rest and is then pulled by a towtruck with a force of 1500N, what is the acceleration of the car?(Ignore frictional losses) The car goes from having a total kinetic energy of 0J tohaving a TKE of 407721J Which means that the tow truck would have to do that amount ofwork to give the car that energy Since work=forcexdistance, we can figure out the distance byusing the energy and force we've been given 407721J=1500Nxdistance means that distance=271.8m Now that we know speed(26.4m/s) and distance(271.8m) we canfind acceleration by saying that final velocity, v=2(distance)/(time) , plugging in ourconstants, we get 26.4m/s=2(271.8m)/time and therefore, since time=20.6s, theacceleration lasted for 20.6s Then we use acceleration, a=v/time, and plug in our constantsto get a=(26.4m/s)/(20.6s)=a=1.28m/s^2
(d) What % error would you make in part (c) if you ignored therotational inertia of the tires and wheels? kinetic energy from rotation(revisited), KE=I^2, wasKE=4x.5(2.8kgm^2)(66rad/s)^2=24394J or KE=24,000J Since the ratio of the rotational intertia to the Total KE is24,000/410,000=.059 or 5.9% I would use 6% errorto have a good round number to work with. The 1100-kg mass of a car includes four tires, each of mass(including wheels) 35kg and diameter 0.80m. Assume each tire andwheel combination acts as a solid cylinder.
Determine
(a) the total kinetic energy of the car when traveling 95km/h
[1] Total kinetic energy = translation + rotation =1/2mv^2+1/2I^2
The velocity converts to v=26.4m/s
I for a solid cylinder= 1/2Mr^2, which for ourcase is 1/2(35kg)(.40m)^2=I=2.8kgm^2
To find for a tire, we need to use(=v/r), =(26.4m/s)/(.40m)=65.97 or just=66rad/s To find the total energy, use eq[1] and plug in all the valueswe just found, since there are four tires, multiply I^2 by4 Total energy=1/2(1100kg)(26.4m/s)^2 +2(2.8kgm^2)(66rad/s)^2=407721Joules or just Totalenergy=410,000J
(b)the fraction of the kinetic energy in the tires and wheels KE=1/2mv^2 There are 4 tires whose combined mass is 4*(35kg)=140kg KEtires=.5(140kg)(26.4m/s)^2=48787J The fraction of the KE in tires = 48787J/407721J=.1197 or 12%or 3/25
(c) If the car is initially at rest and is then pulled by a towtruck with a force of 1500N, what is the acceleration of the car?(Ignore frictional losses) The car goes from having a total kinetic energy of 0J tohaving a TKE of 407721J Which means that the tow truck would have to do that amount ofwork to give the car that energy Since work=forcexdistance, we can figure out the distance byusing the energy and force we've been given 407721J=1500Nxdistance means that distance=271.8m Now that we know speed(26.4m/s) and distance(271.8m) we canfind acceleration by saying that final velocity, v=2(distance)/(time) , plugging in ourconstants, we get 26.4m/s=2(271.8m)/time and therefore, since time=20.6s, theacceleration lasted for 20.6s Then we use acceleration, a=v/time, and plug in our constantsto get a=(26.4m/s)/(20.6s)=a=1.28m/s^2
(d) What % error would you make in part (c) if you ignored therotational inertia of the tires and wheels? kinetic energy from rotation(revisited), KE=I^2, wasKE=4x.5(2.8kgm^2)(66rad/s)^2=24394J or KE=24,000J Since the ratio of the rotational intertia to the Total KE is24,000/410,000=.059 or 5.9% I would use 6% errorto have a good round number to work with.
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