Wheel and Axle
Course (s)B58EC, B58ED–Mech. Eng. Sci. 3, 4 | Year2 | SubjectDynamics | Semester (s)1 and 2 | Laboratory TitleWheel and Axle Acceleration |
Objective: To predict the time taken for a wheel to roll on its axle, down a slope using the energy method.
Theory: Release
Energy Method
Ød or radius r
After descent
m kg
I kg.m2 h v m/sec
ØD or radius R
Figure 1. Energy in a rolling wheel
Referring to Figure 1 when the wheel is released from rest and subsequently rolls down the slope, it accelerates and hence gains energy. Now for a rolling wheel the kinetic energy has two components, translational due to the bodily movement of the mass centre down the slope and rotational due to the wheel spin. Now the source of this energy is the loss in potential energy as the wheel moves down the slope. If it is reasonable to assume that friction effects are insignificant then no energy is lost. Thus the loss in potential energy becomes a gain in kinetic energy.
Hence,
Loss in potential energy = mgh, is equal to the (1)
Gain in kinetic energy = 0.5mv2 + 0.5I 2 (2)
where v = velocity of the mass centre down slope (m/sec) = angular velocity of wheel (rad/sec) = v/r, r is the axle radius when rolling I = Polar moment of inertia = mR2/2
Applying conservation of energy, equate equations 1 and 2 to derive an expression for the velocity v at the bottom of the slope. Using the linear equations of motion, find the expression for time t. Show these derivations in your report.
Experiment:
Using the measured distances (100mm to 500mm, intervals of 100mm) travelled by the wheel and the expressions i.e. (1) velocity at bottom of slope and (2) acceleration down the slope, calculate the time taken for the wheel to roll down the slope. Compare the calculated values with the experimental data
Discussions:
Plot a graph of time t2 vs distance s for calculated and experimental data. Explain the discrepancies between calculated values
Objective: To predict the time taken for a wheel to roll on its axle, down a slope using the energy method.
Theory: Release
Energy Method
Ød or radius r
After descent
m kg
I kg.m2 h v m/sec
ØD or radius R
Figure 1. Energy in a rolling wheel
Referring to Figure 1 when the wheel is released from rest and subsequently rolls down the slope, it accelerates and hence gains energy. Now for a rolling wheel the kinetic energy has two components, translational due to the bodily movement of the mass centre down the slope and rotational due to the wheel spin. Now the source of this energy is the loss in potential energy as the wheel moves down the slope. If it is reasonable to assume that friction effects are insignificant then no energy is lost. Thus the loss in potential energy becomes a gain in kinetic energy.
Hence,
Loss in potential energy = mgh, is equal to the (1)
Gain in kinetic energy = 0.5mv2 + 0.5I 2 (2)
where v = velocity of the mass centre down slope (m/sec) = angular velocity of wheel (rad/sec) = v/r, r is the axle radius when rolling I = Polar moment of inertia = mR2/2
Applying conservation of energy, equate equations 1 and 2 to derive an expression for the velocity v at the bottom of the slope. Using the linear equations of motion, find the expression for time t. Show these derivations in your report.
Experiment:
Using the measured distances (100mm to 500mm, intervals of 100mm) travelled by the wheel and the expressions i.e. (1) velocity at bottom of slope and (2) acceleration down the slope, calculate the time taken for the wheel to roll down the slope. Compare the calculated values with the experimental data
Discussions:
Plot a graph of time t2 vs distance s for calculated and experimental data. Explain the discrepancies between calculated values