A bike with 68.5 cm diameter wheels is traveling on the straight and level down
a paved street. It is moving at 10 m/s. Neglecting friction with the rod and in the bearings,
what frictional force will have to applied by the brake pads to the rims of the wheels to get
the bike to stop within 15 meters. The bike and rider together have a mass of 75 kg.
a paved street. It is moving at 10 m/s. Neglecting friction with the rod and in the bearings,
what frictional force will have to applied by the brake pads to the rims of the wheels to get
the bike to stop within 15 meters. The bike and rider together have a mass of 75 kg.
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A bike with 68.5 cm diameter wheels is traveling on the straight and level down a paved street. It is moving at 10 m/s. Neglecting friction with the rod and in the bearings, what frictional force will have to applied by the brake pads to the rims of the wheels to get the bike to stop within 15 meters. The bike and rider together have a mass of 75 kg.
The friction force produces a torque on the wheel. The work done by the torque reduces the kinetic energy to 0 J.
Initial KE of bike and rider = ½ * 75 * 10^2 = 3750 Joules
Torque = friction force * radius of wheel = Friction force * 0.3425 m
Work = Torque * θ.
θ is the angle, in radians, that the wheel rotates as the bike moves 15 meters.
Work = Friction force * 0.3425 * θ
As the wheel rotates 1 revolution, the bike moves the distance equal to the circumference of the circle.
The number of revolutions that the wheel rotates = 15 m ÷ Circumference
Circumference = 2 * π * 0.3425
Number of revolutions = 15 ÷ (2 * π * 0.3425)
1 revolution = 2 * π radians
Number of radians = Number of revolutions * (2 * π)
Number of radians = [15 ÷ (2 * π * 0.3425)] * (2 * π)
Number of radians = 15 ÷ 0.3425 = θ
Work = Friction force * 0.3425 * (15 ÷ 0.3425)
Work = Friction force * 15
WOW! The radius cancelled.
I did not need to determine the torque or angle. The friction force was applied to the wheel as the tire rolled a linear distance of 15 meters.
Work = Force * linear distance = Friction force * 15
Initial KE of bike and rider = ½ * 75 * 10^2 = 3750 Joules
Friction force * 15 = 3750
Friction force = 3750 ÷ 15 = 250 N
The friction force produces a torque on the wheel. The work done by the torque reduces the kinetic energy to 0 J.
Initial KE of bike and rider = ½ * 75 * 10^2 = 3750 Joules
Torque = friction force * radius of wheel = Friction force * 0.3425 m
Work = Torque * θ.
θ is the angle, in radians, that the wheel rotates as the bike moves 15 meters.
Work = Friction force * 0.3425 * θ
As the wheel rotates 1 revolution, the bike moves the distance equal to the circumference of the circle.
The number of revolutions that the wheel rotates = 15 m ÷ Circumference
Circumference = 2 * π * 0.3425
Number of revolutions = 15 ÷ (2 * π * 0.3425)
1 revolution = 2 * π radians
Number of radians = Number of revolutions * (2 * π)
Number of radians = [15 ÷ (2 * π * 0.3425)] * (2 * π)
Number of radians = 15 ÷ 0.3425 = θ
Work = Friction force * 0.3425 * (15 ÷ 0.3425)
Work = Friction force * 15
WOW! The radius cancelled.
I did not need to determine the torque or angle. The friction force was applied to the wheel as the tire rolled a linear distance of 15 meters.
Work = Force * linear distance = Friction force * 15
Initial KE of bike and rider = ½ * 75 * 10^2 = 3750 Joules
Friction force * 15 = 3750
Friction force = 3750 ÷ 15 = 250 N
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Force arm = (68.5/2) = 34.25cm., = 0.3425m.
Acceleration required = (v^2/2d) = 3.33m/sec^2.
Force = (ma), = 75 x 3.33, = 249.75N. (if applied at centre of wheel).
(249.75 x 0.3425) = 85.54N. at the rim.
Acceleration required = (v^2/2d) = 3.33m/sec^2.
Force = (ma), = 75 x 3.33, = 249.75N. (if applied at centre of wheel).
(249.75 x 0.3425) = 85.54N. at the rim.