(a) Without the wheels, a bicycle frame has a mass of 9.41 kg. Each of the wheel
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Question
(a) Without the wheels, a bicycle frame has a mass of 9.41 kg. Each of the wheels can be roughly modeled asa uniform solid disk with a mass of 0.820 kg and a radius of 0.343m. Find the kinetic energy of the whole bicycle when it is movingforward at 3.25 m/s.1 J
(b) Before the invention of a wheel turning on a axle, ancientpeople moved heavy loads by placing rollers under them. (Modernpeople use rollers, too: Any hardware store will sell you a rollerbearing for a lazy Susan.) A stone block of mass 941 kg moves forward at 0.325m/s, supported by two uniform cylindrical tree trunks, each of mass82.0 kg and radius 0.343 m. There is no slipping between the blockand the rollers or between the rollers and the ground. Find thetotal kinetic energy of the moving objects.
2 J
Explanation / Answer
There are two types of motion: linear and rotational. You mustconsider both to find the total kinetic energy. (a) We need to find the angular velocity, , of the bicyclewheels before we can find the energy due to their rotation. If thewhole bicycle is moving at 3.25 m/s, we need to find out how manytimes a second the wheel rotates. The circumference of a circle is2r, so for a wheel with radius 0.343, we get C = 2r = 2 * * 0.343 m = 2.155 m So, one rotation of the wheel will take the bicycle 2.155meters. Divide the linear velocity by the circumference to get the numberof rotations per second: v / C = 3.25 (m/s) / 2.155 m/rotation = 1.51 rotation/s There are 2 radians in one full rotation, so for weget = 1.51 rotation/s * 2 rad/rotation = 9.48 rad/s Now we can find the total rotational energy of the wheels: K = 1/2I2 I, the moment of inertia, for a solid disk (which we can assume thewheels are) is 1/2mr2 around an central axisperpendicular to the surface (which is the axis a wheel rotatesabout). I = 1/2 * 0.820 kg * 0.3432 m2 = 0.0482kg-m2 So, the energy for one wheel: K = 1/2 * 0.0482 * 9.482 = 2.16 J So the energy for the rotation of both wheels is about4.33 J. Add in the bike plus the wheels (you need to consider that thewheels are moving forward with the bike!), using K =1/2mv2: K = 1/2 * (2 * 0.820 + 9.41 kg) * 3.252m2/s2 + 4.33 J = 62.7 J (b) We need to do exactly the same thing here. A cylinder rotatinglike a wheel behaves just like a disk would. = v / r (this is a more efficient method; the last time Iworked through the 2 factor so that you can see where itcomes from) = 0.325 (m/s) / 0.343 m = 0.95 rad/s EDIT 2: Caught another mistake. The logsweigh 82 kg each. I = 1/2 * 82 kg *0.3432 m2 = 4.82 kg-m2 K = 1/2I2 = 1/2 * 4.82 kg-m2 * 0.952rad/s = 2.175 J Since there are two, the total kinetic energy for the rotation ofthe logs is 4.35 J. EDIT: I made a mistake here. Since thelogs are rollers, they do not move with theblock. Again, add in the linear energy to find the total using K =1/2mv2: 1/2 * (941) kg * 0.3252 m2/s2 +4.35 J =54 J [This should now becorrect]
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