Question

a) Without the wheels, a bicycle frame has a mass of 8.75 kg. Each of the wheels can be roughly modeled as a uniform solid disk with a mass of 0.820 kg and a radius of 0.343 m. Find the kinetic energy of the whole bicycle when it is moving forward at 3.00 m/s. J (b) Before the invention of a wheel turning on an axle, ancient people moved heavy loads by placing rollers under them. (Modern people use rollers, too. Any hardware store will sell you a roller bearing for a lazy susan.) A stone block of mass 875 kg moves forward at 0.300 m/s, supported by two uniform cylindrical tree trunks, each of mass 82.0 kg and radius 0.343 m. No slipping occurs between the block and the rollers or between the rollers and the ground. Find the total kinetic energy of the moving objects. J

Answer 1

Answer:

a) The total kinetic energy = 50.445 Joules

b) Total kinetic energy = 42.14 Joules

Explanation:

Mass of frame, $m_{f} = 8.75 kg$

Mas of wheel, $m_{w} = 0.820 kg$

Radius of the wheel, $R = 0.343 m$

a) The motion undergone by the bicycle is both translational and rotational

KE = Kinetic Energy

The translational kinetic enegy = KE due to the frame + KE due to the first wheel + KE due to the second wheel

KE due to the frame, $KE_{f} = 0.5 m_{f} v^{2}$

KE energy due to the first and second wheels, $KE_{w} = 0.5m_{w} v^{2} + 0.5m_{w} v^{2}$

$KE_{w} = m_{w} v^{2}$

The translational Kinetic energy = $0.5 m_{f} v^{2} + m_{w} v^{2}$

The rotational kinetic energy = sum of the rotational energies of the two wheels

Rotational kinetic energy = $0.5I_{w} w^{2} + 0.5I_{w} w^{2}$

Rotational kinetic energy = $I_{w} w^{2}$

$I_{w} = 0.5m_{w} R^{2} \\w = \frac{v}{R}$

Speed, v = 3.00 m/s

Rotational kinetic energy = $0.5m_{w} R^{2} (\frac{v}{R}) ^{2}$

Rotational kinetic energy = $0.5m_{w} v^{2}$

The total kinetic energy = $0.5 m_{f} v^{2} + m_{w} v^{2} + 0.5m_{w} v^{2}$

The total kinetic energy = $v^{2} (0.5m_{f} + 1.5m_{w} )$

The total kinetic energy = $3^{2} (0.5*8.75 + 1.5*0.820)$

The total kinetic energy = 50.445 Joules

b) Mass of stone $m_{st} = 875 kg$

Mass of tree trunks, $m_{tr} = 82.0 kg$

Radius, R = 0.343 m

Velocity, v = 0.3 m/s

Total translational kinetic energy due to the stone and the tree trunk

translational kinetic energy = $0.5m_{st} v^{2}+ 0.5m_{tr} v^{2} + 0.5m_{tr} v^{2}$

The speed of the tree trunk will be half that of the stone

Translational kinetic energy = $0.5m_{st} v^{2}+ m_{tr} (\frac{v}{2}) ^{2}$

Translational kinetic energy = $0.5m_{st} v^{2}+ 0.25 m_{tr} v^{2}$

Rotational kinetic energy = $0.5m_{tr} R^{2} (\frac{v}{R}) ^{2}$

since speed = v/2

Rotational kinetic energy = $0.5m_{tr} R^{2} (\frac{v}{2R}) ^{2}$

Rotational kinetic energy = $0.125m_{tr} v^{2}$

Total kinetic energy = $0.5m_{st} v^{2}+ 0.25 m_{tr} v^{2} + 0.125 m_{tr} v^{2}$

Total kinetic energy = $0.5m_{st} v^{2}+ 0.375 m_{tr} v^{2}$

Total kinetic energy = $(0.5m_{st}+ 0.375 m_{tr}) v^{2}$

Total kinetic energy = $(0.5*875+ 0.375*82) 0.3^{2}$

Total kinetic energy = 42.14 Joules