Physics 1307 Practice Quiz 5 Chapter 8 1/3
Physics 1307 Practice Quiz 5 Chapter 8 Problem I : Satellite maneuvering In order to get a flat uniform cylindrical satellite spinning at the correct rate, engineers fire four tangential rockets as shown. If the satellite has a mass of M=2600Kg a radius of 3.0 m. What is the required force of each rocket if the satellite is to reach 30 rmp in 5.0 min? Problem II: Atwood's machine v 2.0 An Atwood's machine consists of two masses m 1 and m2 , which are connected by a massless inelastic cord
that passes over a pulley. If the pulley has radius R and moment of inertia I about its axle, determine the acceleration of the masses m 1 and m2 and compare to the situation in which the moment of inertia of the pulley is ignored. (Hint: The tensions are not necessarily equal)
SOLUTIONS Problem solving strategy Momentum conservation and collisions 1. As always, draw a clear and complete diagram. 2. Draw a free body diagram for the body under consideration (or for each body if more than one), showing only (and all ) the forces acting on that body and exactly where they act, so you can determine the torque due to each. 3. Identify the axis of rotation and calculate the torques about it. Choose positive and negative directions of rotation (counterclockwise and clockwise), and assign the correct sign to each torque. 4. Apply Newton's second law for rotation, ∑ =I . If the moment of inertia is not given, and it is not the unknown sought, you need to determine it first. Use consistent units, which in SI are: in rad / s2 ; in m.N and I in Kg⋅m2 . You have one equation for each component. 5. Solve algebraically for the unknown quantity (ies).
1/3
Problem I First, a graph of the situation and then the application of Newton's second law for rotation:
∑ =I
4FR=I 1 MR 2 I 2 1 F= = = M R 4R 4R 8 1min rev 30 2 1 min 60s F= 2600Kg3.0 m =10 N 8 s 5.0 min 60 min
R
Problem II We start with a picture of the system and the free body diagrams for each object.
R
+
m1 m2 FT2
FT1 mm2 2
m1
m2g
m1g F T2−m2 g=m2 a 2
F T1−m 1 g=m 1 a1
2/3
∑ =I
−F T2 RFT1 R=I
FT2
FT1
Now, we can observe that we have only 3 equations and 5 unknowns: F T1 , F T2 , , a1 , a 2 since we are not assuming that the tensions F T1 , F T2 are equal. We need to find two more equations. The first one is easy:
a 1=−a 2 since the masses are linked by a massless inelastic cord. The second equation uses the fact that the acceleration of either block and the angular acceleration of the pulley are also linked. For example we have =
−a 1 R
Inserting the two equations for the accelerations and the equations for the two tensions into the equation for the pulley we find: ∑ =I −a 1 −m 2 −a1 m2 g Rm1 a 1m 1 g R=I R −a 1 m1m2 a 1 m1−m2 g=I 2 R I a1 m2m1 2 = m2−m1 g R m2−m1 g a1 = I m 2m1 2 R If we compare this result with the one that we obtained ignoring the moment of inertia m −m1 g a 1= 2 we realize that they are the same if we take I=0, which is to be expected. m1m2 3/3
that passes over a pulley. If the pulley has radius R and moment of inertia I about its axle, determine the acceleration of the masses m 1 and m2 and compare to the situation in which the moment of inertia of the pulley is ignored. (Hint: The tensions are not necessarily equal)
SOLUTIONS Problem solving strategy Momentum conservation and collisions 1. As always, draw a clear and complete diagram. 2. Draw a free body diagram for the body under consideration (or for each body if more than one), showing only (and all ) the forces acting on that body and exactly where they act, so you can determine the torque due to each. 3. Identify the axis of rotation and calculate the torques about it. Choose positive and negative directions of rotation (counterclockwise and clockwise), and assign the correct sign to each torque. 4. Apply Newton's second law for rotation, ∑ =I . If the moment of inertia is not given, and it is not the unknown sought, you need to determine it first. Use consistent units, which in SI are: in rad / s2 ; in m.N and I in Kg⋅m2 . You have one equation for each component. 5. Solve algebraically for the unknown quantity (ies).
1/3
Problem I First, a graph of the situation and then the application of Newton's second law for rotation:
∑ =I
4FR=I 1 MR 2 I 2 1 F= = = M R 4R 4R 8 1min rev 30 2 1 min 60s F= 2600Kg3.0 m =10 N 8 s 5.0 min 60 min
R
Problem II We start with a picture of the system and the free body diagrams for each object.
R
+
m1 m2 FT2
FT1 mm2 2
m1
m2g
m1g F T2−m2 g=m2 a 2
F T1−m 1 g=m 1 a1
2/3
∑ =I
−F T2 RFT1 R=I
FT2
FT1
Now, we can observe that we have only 3 equations and 5 unknowns: F T1 , F T2 , , a1 , a 2 since we are not assuming that the tensions F T1 , F T2 are equal. We need to find two more equations. The first one is easy:
a 1=−a 2 since the masses are linked by a massless inelastic cord. The second equation uses the fact that the acceleration of either block and the angular acceleration of the pulley are also linked. For example we have =
−a 1 R
Inserting the two equations for the accelerations and the equations for the two tensions into the equation for the pulley we find: ∑ =I −a 1 −m 2 −a1 m2 g Rm1 a 1m 1 g R=I R −a 1 m1m2 a 1 m1−m2 g=I 2 R I a1 m2m1 2 = m2−m1 g R m2−m1 g a1 = I m 2m1 2 R If we compare this result with the one that we obtained ignoring the moment of inertia m −m1 g a 1= 2 we realize that they are the same if we take I=0, which is to be expected. m1m2 3/3
