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Get the assessment: FA08.2.docx
Get more Practice Problems: Worksheet-AngularDynamics8.2.docx
The assessment solved:
I (about centers): cylinder = 1/2mr2, ring/point = mr2, sphere = 2/5mr2, rod = 1/12mL2 (= 1/3mL2 about end)
1. A mechanic needs to exert 385 mN of torque. He weighs 833 N and he stands on the handle of his wrench that is making a 17.0o angle above the horizontal. How far from the center must he stand? (Be careful what you use for the angle)
(0.483 m)
2. What is the acceleration of a flywheel with a moment of inertia of 0.145 kg m2 if a torque of 2.80 mN acts on it?
(19.3 rad/s/s)
More like this: (From the practice problems)
3. A 0.680 m diameter flywheel has a moment of inertia of 0.243 kg m2. What is the angular acceleration of the flywheel if you exert 4.50 N tangentially at the edge to speed it up?
(6.29 rad/s/s)
More like this: (From the practice problems)
4. A 0.210 m radius grinding disk is spinning at 1350 RPM. If it goes through 85.0 rotations being brought to rest by a 1.20 N frictional force applied tangentially at its edge, what is the moment of inertia of the disk?
(0.0135 kg m2)
More like this: (From the practice problems)
5. A 4.30 m diameter (cylindrical) merry go round going 45.0 RPM stops in 37.0 rotations because of an 8.30 N force applied tangentially at the edge. What is the mass of the merry go round?
(162 kg)
More: (From the practice problems)
More Practice Problems: Worksheet-AngularDynamics8.2.docx
Angular Dynamics problems from 8.2 - Links are video solutions!!!!!!
A. What force acting at 25.0o with a line perpendicular to the end of a 13.0 cm long wrench will generate 7.80 mN of torque about the left side of the wrench? (66.2 N) FA8 2 #1 Ex1
B. Calculate the torque about the left side of the wrench if 52.0 N acts at an 21.0o angle with the end of a 13.0 cm long wrench. (2.42 mN) FA8 2 #1 Ex2
C. A 35.0 cm wrench makes a 23.0o angle above the horizontal. What is the torque about the left side of the wrench if a 24.0 N force is exerted vertically upward at the end? (7.73 mN) FA8 2 #1 Ex3
D. A force is exerted at an angle of 129o with a 16.0 cm wrench as shown below. Calculate the force needed to create 3.80 mN of torque about the left side of the wrench. (30.6 N) FA8 2 #1 Ex4
Moments of inertia: Cylinder: I = 1/2 mr2, Sphere: I = 2/5mr2, Thin Ring or Point Mass: I = mr2
Moments of inertia: Cylinder: I = 1/2 mr2, Sphere: I = 2/5mr2, Thin Ring or Point Mass: I = mr2
Simple F = ma problems:
1. A baton requires 5.70 mN of torque to accelerate at 18.4 rad/s/s about its center. What is the moment of inertia? (0.310 kgm2) Rotational Dynamics Ex #1
2. A flywheel with a moment of inertia of 0.859 kg m2 accelerates at 13.0 rad/s/s. What is the torque? (11.2 mN) Rotational Dynamics Ex #2
3. A motor with 43.0 mN of torque accelerates at 153 rad/s/s. What is its moment of inertia? (0.281 kgm2) Rotational Dynamics Ex #3
4. A torque of 21.0 mN acts on a motor with a moment of inertia of 1.53 kg m2. What is the angular acceleration? (13.7 rad/s/s) Rotational Dynamics Ex #4
5. What torque will accelerate a motor with a moment of inertia of 3.87 kg m2 at 6.60 rad/s/s? (25.5 mN) Rotational Dynamics Ex #5
F = ma problems, but I = 1/2 mr2 (cylinder), 2/5mr2 (sphere), or kinematics, or
6. A 0.400 m diameter, 4.30 kg sphere accelerates about its center at 6.80 rad/s/s. What is the torque? (0.468 mN) Rotational Dynamics Ex #6
7. A drill with a moment of inertia of 0.0180 kg m2 is slowed by a frictional torque of 0.270 mN. If it is moving at 142 rad/s, how many radians will it go through before it stops? (672 rad) Rotational Dynamics Ex #7
8. A grinding wheel with a diameter of 0.640 m and a moment of inertia of 0.172 kg m2 decelerates at -8.90 rad/s/s because of a tangential friction force applied at the edge. What is this force? (4.78 N) Rotational Dynamics Ex #8
9. A torque of 19.0 mN acts on a flywheel with a moment of inertia of 3.20 kg m2. If it starts at rest, in what time will it go through 16.0 radians? (2.32 s) Rotational Dynamics Ex #9
10. A torque of 3.50 mN acts on a 7.10 kg, 0.132 m diameter shot put. (a sphere) What is the angular acceleration of the sphere? (283 rad/s/s) Rotational Dynamics Ex #10
Same as above with unit conversions:
11. A 0.219 m diameter bowling ball has a tangential force 5.50 N acting on it and it accelerates from rest going through 13.0 rotations in 3.21 seconds. What is the moment of inertia of the ball? (0.0380 kgm2) Rotational Dynamics Ex #11
12. A 0.310 m radius flywheel (essentially a thin ring) with a mass of 3.20 kg. What is its rate of deceleration if you exert a force of 2.20 N tangentially at its edge? (2.22 rad/s/s) Rotational Dynamics Ex #12
13. A flywheel is a 13.2 kg 1.80 m diameter thin ring. If you exert a force of 51.0 N tangentially at its edge, what is its angular acceleration? (4.29 rad/s/s) Rotational Dynamics Ex #13
14. A flywheel that is a 0.730 m diameter thin ring with a mass of 16.0 kg would require what torque to accelerate from rest to 1120 RPM in 8.10 seconds? (30.9 mN) Rotational Dynamics Ex #14
15. What is the moment of inertia of a 0.258 m radius flywheel if when you exert a tangential force of 11.5 N at the edge it accelerates from rest to 680. RPMs in 123 rotations? (0.904 kgm2) Rotational Dynamics Ex #15
Same as above with unit conversions and kinematics:
16. A 161 kg 4.72 m diameter (cylindrical) merry go round is sped up from rest by a 25.0 N force applied tangentially at its edge. What is its speed in RPMs after 38.0 seconds? (47.8 RPM) Rotational Dynamics Ex #16
17. A 2.10 m radius, 351 kg (cylindrical) merry go round spinning at 75.0 RPM slows to a halt in 11.5 rotations. What force applied tangentially at the edge would cause this? (157 N) Rotational Dynamics Ex #17
18. A 232 kg 4.10 m diameter (cylindrical) Merry go round is stopped from a speed of 94.0 RPM in 55.0 seconds. What frictional force applied tangentially at the edge would cause this? (42.6 N) Rotational Dynamics Ex #18
19. A 243 kg 1.70 m radius (cylindrical) merry go round stops from a speed of 68.0 RPM because of a frictional force applied at the edge of 8.50 N. How many rotations does it go through in stopping? (98.1 rotations) Rotational Dynamics Ex #19
20. A 4.60 m diameter (cylindrical) merry go round speeds up from rest going through 5.10 rotations in 41.0 seconds because of a 15.0 N force applied tangentially at the edge. What is the mass of the merry go round? (342 kg) Rotational Dynamics Ex #20
The instruction for this unit:
Conversions and tangential relationships:
2:08 Basic Quantities
Angle Conversions:
Angle Conversion #1 (rot to rad)
Angle Conversion #2 (rad to rot)
Angle Conversion #3 (deg to rad)
Angle Conversion #4 (rad to deg)
Velocity Conversions:
1:20 Anglular Velocity Conversions
Angular Velocity Conversion #1 (rot/s to rad/s)
Angular Velocity Conversion #2 (rad/s to rot/s)
Angular Velocity Conversion #3 (RPM to rad/s)
Angular Velocity Conversion #4 (rad/s to RPM)
Angular Velocity Conversion #5 (rot/s to RPM)
Angular Velocity Conversion #6 (RPM to rot/s)
Distance and angle:
Tangential Relationships #1 (find s)
Tangential Relationships #2 (find angle)
Velocity and angular velocity (omega):
Tangential Relationships #3 (find v)
Tangential Relationships #4 (find omega)
Acceleration and angular acceleration (alpha):
Tangential Relationships #5 (find a)
Tangential Relationships #6 (find alpha)
Tangential with unit conversions:
Tangential Relationships #7 (rot, r to find s)
Tangential Relationships #8 (s, r to find rot)
Tangential Relationships #9 (RPM, r to find v)
Tangential Relationships #10 (v, r to find RPM)
Angular Dynamics
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