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HINTS FOR PROBLEM SET #14
Problem Set #14 Problem 1 This problem is one of a rigid body rotating about a fixed axis. It is completely analogous to the problem of a particle moving in rectilinear motion. Instead of the one position coordinate, x, there is just the one angular coordinate, q. In both situations the position of the body is completely determined by one coordinate, x or q. Hence all the equations are merely scalar equations.
Part a) You are asked to find the number of revolutions (an angular displacement) that a shaft will execute before coming to rest. So you need to find q when the value of w is zero. You are given the relationship of a (angular acceleration) as a function of w (angular velocity).
Use the kinematic equation: a = dw/dt where a = - 0.25 w So: - 0.25 w = dw/dt Now substitute dq/dt for w: - 0.25 dq/dt = dw/dt - 0.25 dq = dw Now integrate. The limits of integration on the dq side will be between 0 and q. The limits of integration on the dw side will be between 20 rad/sec and 0.
Solve for the upper limit of q. You should get q = 80 radians = 12.72 revolutions.
Part b) You need to find the time required for the shaft to come to rest. So you need w as a function of t. Use the kinematic equation: a = dw/ dt where a = - 0.25 w So: - 0.25 w = dw/dt Separate the variables: - 0.25 dt = dw/w Now integrate. The limits of integration on the dt side will be between 0 and t. The limits of integration on the dw side will be between 20 rad/sec and 0.
Solve for the upper limit of time. You should get t = ₯.
Part c) This is essentially the same problem as part (b). The only difference is the upper limit on the integral for the w side of the equation. Instead of 0, it will now be 1% of the initial value of w (= 0.01 * 20 rad/sec). You should get t = 18.42 sec.
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Problem Set #14 Problem 2 The assembly shown is a rigid body rotating about a fixed axis AB. To find the velocity and acceleration of a point, E, on the rigid body, just plug into the kinematic equations for the velocity and the acceleration of a point on a rigid body that is rotating about a fixed axis: vE = w x rE/P aE = [a x rE/P ] + [w x (w x rE/P )] where w = w l a = al l is the unit vector in the direction of the axis of rotation P is any point on the axis of rotation rE/P = the position vector from any point on the axis of rotation to the point E
For this particular problem, the axis of rotation goes through the points A and B. So you need to find the unit vector, l, in the direction from A to B. Once you find this unit vector you can get expressions for w and a: w = (7.5 rad/sec)l a = 0l (because the angular velocity is constant)
For the position vector, rE/P, you can use either point A or B for point P. If you use point B than rE/P = rE/B and is simply -12k.
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Problem Set #14 Problem 3 Consider two meshed gears that are rotating. Call the point of contact on one gear, point C and the point of contact on the other gear, point D.
Part a) If a belt is moving over a pulley without slipping, the points of contact between the belt and the pulley are like two meshed gears. In the picture shown for this problem, consider B to be a point on the belt, not the pulley. Let point P be the corresponding point on the pulley that is in contact with point B.
Assume that the belt is inextensible. Hence: vA = vB (directions will be different) aA = aB (directions will be different)
From #1 above: vB = vP Hence vP = vA
From #2 above: aA = aPt
Using the kinematic (scalar) equations for point P: vP = wr = vA aPt = ar = aA
vA, aA and r are given. You can solve for w and a
Part b) To find the total acceleration of the point P on the pulley you need to find the normal component of acceleration and then add it (using vector addition) to the tangential component that you found in part (a) of this problem.
The magnitude of the normal component of acceleration is equal to v2/r = (wr)2/r = w2r The direction of the normal component of acceleration points towards the center of the pulley.
Adding the two components (tangential and normal) of the acceleration of point B will give a vector of magnitude 38.56 in/s2 in a direction 76.5o below the horizontal and pointing to the right. (Since the tangential acceleration is negative and points towards the right.)
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Problem Set #14 Problem 4 Part a) The two cylinders that block A and block B are wrapped around are rigidly connected. They rotate together.
Recall the kinematic equation for uniformly accelerated rectilinear motion: x = x0 + v0t + ½at2 Use the analogous kinematic equation for uniformly accelerated rotation: q = q0 + w0t + ½at2 where q0 = 0 w0= vA/rlarge = 0.6 rad/s a = aAt/rlarge = 0.375 rad/s2 t = 6 seconds This will enable you to find the number of revolutions executed by the pulley in 6 sec.
Part b) Recall the kinematic equation for uniformly accelerated rectilinear motion: x = x0 + at Use the analogous kinematic equation for uniformly accelerated rotation: w= w0 + at to find w after 6 seconds. Once you know the angular velocity, w, of the pulley, you can find the velocity of point P on the pulley which is in contact with the rope attached to block B: vP = w rsmall This is equal to the velocity of point B, vB.
The position of B can be determined by knowing how much rope has wrapped around the inside diameter. This is equal to the angular displacement (found in part a) times the radius of the pulley.
Part c) To find the acceleration of point C on the rim, you need to find both the tangential and the normal components of point C and then add these two vectors together. aCt = a rlarge aCn = v2/rlarge = (wrlarge)2/rlarge = w2rlarge aC = sqrt (aCt2 + aCn2) at an angle of tan-1(aCn/aCt)
This should give you a vector of magnitude 104 mm/s2 with a direction of 43.8 degrees below a horizontal line and pointed to the left. |