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Discussion

The animation shown above demonstrates the kinematics of the motion of the bar/disk system. The bar rotates about end O. The disk is pinned to end A of the bar, and rolls without slipping on the fixed circular surface. Since the disk rolls without slipping, when point C on the outer circumference of the disk is in contact with fixed ground, the velocity of C is zero. You can readily see this in the animation above.

The instant center (IC) for the disk is the contact point of the disk with the fixed surface. Note the relative sizes of the speeds of A and C in comparison to their distances from the IC. When are the speeds the same? At what positions is the speed of C twice that of the speed of A?

You are asked to determine the rotation rate of the bar when the bar reaches the vertical position shown below.

HINT: As always, we should follow the four-step plan for solving this problem.
STEP 1: FBD. We will be using the work/energy equation to determine this angular speed. Based on earlier recommendations, we will make the choice of our system BIG, including the bar, and disk together.
STEP 2: Kinetics (here, work/energy). The total kinetic energy of the system shown in your FBD above is that of the bar + that of the disk. For the bar, it is recommended that you choose the fixed point O as your reference point for the KE. For the disk, it is recommended that you choose the center of mass A.  Be sure to identify the datum line for the gravitational potential energy, and use this in writing down this potential.
STEP 3: Kinematics. You need to relate the speed of A to the angular speeds of the bar and of the disk.
STEP 4: Solve

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Discussion

The animation shown above demonstrates the kinematics of the motion of the system made up of bars OA and AB. Bat OA rotates about the fixed point O. Bar AB is attached of OA with a pin at A, and end B is constrained to move along in a smooth, horizontal slot.

You are asked to determine the speed of end B of AB when the bar OA reaches the vertical position shown below.

QUESTION: Where is the instant center (IC) of AB for the position above? What are the consequences of this location for IC_AB? Can you see these  in the above image when you consider the relative sizes of the speeds of points A, G and B?

HINT: As always, we should follow the four-step plan for solving this problem.
STEP 1: FBD. We will be using the work/energy equation to determine this angular speed. Based on earlier recommendations, we will make the choice of our system BIG, including both bars together.
STEP 2: Kinetics (here, work/energy). The total kinetic energy of the system shown in your FBD above is that of the bar OA + that of bar AB. For bar OA, it is recommended that you choose the fixed point O as your reference point for the KE. For bar AB, it is recommended that you choose the center of mass G for the reference point for its KE.  Be sure to identify the datum line for the gravitational potential energy, and use this in writing down this potential. What is the work done by force F?
STEP 3: Kinematics. Locate the IC for bar AB. What does this say about the angular speed of AB? Use this result in solving your energy equation.
STEP 4: Solve

Ask your questions here. Or, answer questions of others here. Either way, you can learn.

DISCUSSION and HINTS

Recall the following four-step plan outline in the lecture book and discussed in lecture:

Step 1: FBDs
Draw a single free body diagram (FBD) of the bar and the particle.

Step 2: Kinetics (Work/energy)

• Looking at your FBD above, which forces, if any, do work that is not a part of the potential energy of the system? If there are none, then energy is conserved.
• Write down the individual contributions to the kinetic energy from the bar and particle, and add together for the total kinetic energy of the system.
• Define your gravitational datum line. Write down the individual contributions to potential energy from the bar, particle and spring, and add together for the total potential energy.

Step 3: Kinematics
Use the following rigid body kinematics equation to relate the angular velocity of the bar to the velocity of the particle B at position 2:

v_B= v_G + ω_AB x r_B/G

Step 4: Solve
From your equations in Steps 2 and 3, solve for the angular velocity of the bar.

Ask your questions here. Or, answer questions of others here. Either way, you can learn.

DISCUSSION and HINTS

The animation below shows the motion of the wheel as it is being pulled up the incline. The animation shows the velocity of a number of points on the wheel. Think about the location of the instant center for the wheel as it rolls without slipping, and how this location affects the direction and magnitude of the velocities shown here.

Recall the following four-step plan outline in the lecture book and discussed in lecture:

Step 1: FBDs
Draw a free body diagram (FBD) of the wheel.

Step 2: Kinetics (Newton/Euler)

• Looking at your FBD above, which forces, if any, do work that is not a part of the potential energy of the system? Pay particular attention to the friction force at the no-slip point on the wheel - does it do work? How do you find the work done by the applied force F?
• Write down the kinetic energy of the wheel. Recall that the expression the KE for the planar motion of a rigid body is: T = 0.5*m*v_A² + 0.5*I_A*ω², where A is either the center of mass or a fixed point (fixed points include instant centers). So, in this case you can use either O or the no-slip contact point (let's call that point C).
• Define your gravitational datum line. Write down the potential energy for the wheel.

Step 3: Kinematics
Use the IC approach to relate the velocity of the point on the wheel where the cable comes off (call that point A) to the angular velocity of the wheel. Or, instead, use the following rigid body kinematics equation:

v_A= v_C + ω x r_A/C

Through a time integration of the relationship between the speed of A and the angular velocity of the wheel, you can determine the distance through which F moves.

Step 4: Solve
From your equations in Steps 2 and 3, solve for the angular velocity of the wheel at position 2.

NOTE: The system is released from rest in a vertical plane.

DISCUSSION

Recall the four-step plan for solving kinetics problems.

Step 1: FBDs
Here it is recommended (as for all problems using the Newton/Euler approach) to draw individual FBDs for the system: one each for block A and block B, and for the pulley.

Step 2: Kinetics (Newton/Euler)
Write down the Euler equation for the pulley, along with the Newton equations for A and B in the vertical direction. With that, you have three equations and five unknowns (T_A, T_B, a_A, a_B and α_pulley). Please note that the tensions T_A and T_B are NOT equal to the weights of blocks A and B.

Recall from ME 270 that friction opposes (impending) motion. For this problem, you need to decide in which direction (CW or CCW) the pulley will be rotating. To do this, consider first the friction-free case. Once the direction of rotation is determined from that, draw the pulley FBD with the friction torque opposing that motion.

Step 3: Kinematics
It is recommended that you write down the acceleration of points C and E on the pulley (see figure above) in terms of the acceleration of the pulley center O. Then, relate the accelerations of A and B to the accelerations of C and E, respectively.

Step 4: Solve
At this point, you have five equations and five unknowns. Solve!

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Discussion and hints:

Recall the following four-step plan outline in the lecture book and discussed in lecture:

Step 1: FBDs
Draw individual free body diagrams for the bar and the disk.

Step 2: Kinetics (Newton/Euler)
Write down the Newton/Euler equations for the disk and for the bar based on your FBDs above.

Step 3: Kinematics
You need to use the rigid body acceleration equations to relate the accelerations of O and the center of mass of the bar, G, through the angular acceleration of the bar. Also, you need to relate the acceleration of O to the angular acceleration of the disk.

Step 4: Solve
Solve your equations for the angular accelerations of the bar and the disk.

Any questions?

NOTE: The view of the figure provided is that of the rod as seen from above the surface on which the rod lies.

Any questions?? Please ask/answer questions regarding this homework problem through the "Leave a Comment" link above.

Discussion and hints:

Recall the following four-step plan outline in the lecture book and discussed in lecture:

Step 1: FBDs
Draw a free body diagram of the bar.

Step 2: Kinetics (Newton/Euler)
Write down the Newton/Euler equations for the bar using your FBD above. Take care in choosing the reference point for your moment equation. In order to use the "short form" of Euler's equation, this point should be either a fixed point or the body's center of mass. For this problem, there are no fixed points.

Step 3: Kinematics
The paths of A and B are known: A travels on a straight path aligned with the inclined wall, and B travels on a circular path centered at O. Since the bar is released from rest, you know that the speeds of A and B are zero - therefore, the centripetal component of acceleration for each point is zero. This leaves the acceleration of points A and B tangent to their paths. (You can see this from the animation above for the instant when AB is horizontal.) It is recommended that you use two kinematics equations: one relating points A and B, and the other relating the center of mass G of the bar to either A or B.

Step 4: Solve
Solve your equations above for the tension in cable BO.

Any questions?

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NOTE: The system moves in a horizontal plane.

DISCUSSION
Using the four-step plan:
STEP 1: Free body diagram (FBD) - Draw an FBD of bar AB.
STEP 2: Kinetics - Write down the Newton/Euler equations for the bar based on your FBD above. For the "short form" of the Euler equation, please note that you are constrained to using a moment about the center of mass G since there are no fixed points on AB; that is, you must use ΣM_G = I_G α.
STEP 3: Kinematics - With the inextensible cable being taut, all points on the rigid body AB have the same acceleration, and the angular acceleration of AB is zero: α = 0.
STEP 4: Solve - Use the equations from STEPS 2 and 3 to solve for the tension in the cable and the reaction at on the bar at A.

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