The first question involves an atwood machine, the second involves two different rods attached to hinges falling/rotating to the ground.
See the jpeg for exact questions. The text is provided below simply for the benefit of the search engine:
A frictionless pulley with mass Mb is attached to the ceiling, in a gravity field g. Mass Mc is smaller than mass Ma. The tensions Tx, Ty, Tz, and the constant g are magnitudes. For each statement, select a response. Motion of Masses on a Pulley.
Greater than Less than Equal to True False Ty is .... Tz +Tx.
Greater than Less than Equal to True False Tz is .... Tx.
Greater than Less than Equal to True False (Ma+Mc+Mb)g is .... Ty.
Greater than Less than Equal to True False The magnitude of the acceleration of Ma is .... that of Mc.
Greater than Less than Equal to True False The center-of-mass of Ma, Mc, and Mb does not accelerate.
Greater than Less than Equal to True False Tz is .... (Mc)g.
Two uniform rods are connected to a table by pivots at one end. Rod B is longer than rod A. Both are released simultaneously from an initial angle q as shown in the figure. Notation: CM = center of mass; a = angular acceleration; ax = magnitude of horizontal acceleration; ay = magnitude of vertical acceleration. For each of the following statements, determine whether it is correct or incorrect.
Correct Incorrect ax of the CM initially equals 0 for both rods.
Correct Incorrect The density of the rods affect their rate of fall.
Correct Incorrect aA and aB are the same initially.
Correct Incorrect Just before landing, the CM of B has a smaller speed than the CM of A.
Correct Incorrect Rods A and B hit the ground at the same time.
Correct Incorrect aA and alphaB both increase with time.
Correct Incorrect aA and aB are dependent on q.
Correct Incorrect ay is initially equal for the CM of A and B.
SOLUTION This solution is FREE courtesy of BrainMass!
Please see the attachment
The accelerations of the masses and the tensions in the strings can be calculated in this way---
As the pulley is frictionless it will apply no friction force between the pulley and the string in tangential direction and for that tension through the string remains the same. Hence Tx = Tz
As the C.M. of the pulley is not moving the net vertical force on the pulley should be zero. Hence
And as Ty = Tx + Tz + Mb
So the responses
1 Greater then
2 Equal to
As the tension in the string Tx = Ty is given by 2MaMc/(Ma + Mc) which is less then Ma and Mc both the response
3 Greater then
As the length of the string is not changing
4 Equal to
As Mc < Ma the acceleration of CM is downwards
As Mc accelerates upwards net force on Mc is upwards
6 Greater then
When a rod is released as given, torque acting due to the gravitational force, which is acting at the CM of the rod. Hence the torque is given by
= force X perpendicular distance from the axis of rotationa
= mg X (L/2) cos = (m g L cos)/2 [L is length of the rod]
Thus the angular acceleration of the rod is given by
= / I
(I is Moment of inertia of the rod about the axis of rotation = (mL2/3)
= (m g L cos /2)/(mL2/3) = (3 g cos)/2L --------equation-A
The acceleration of CM of the rod is a = L/2 = (3 g cos)/4
Which is perpendicular to the rod. Hence
ax = a sin and ay = a cos
= (d/d) = (3 g cos)/2L integrating this we get
angular velocity of the rod just before impect with ground is
= [(3gL sin/2]
1 incorrect As both rods starts accelerating as soon as released.
2. incorrect As the angular acceleration does not depend on mass
3 incorrect As the angular acceleration depends on the length
4. incorrect As angular velocity and speed both increases with the
5 Incorrect As the angular acceleration is inversely proportional to L
6 correct As with time decreases and cos increases
7 correct As shown by formula
8 correct As cos is same for both at that instance and is
independent of L