Tag Archives: Rotational Dynamics

A2L Item 187

Goal: Problem solving

Source: UMPERG-ctqpe160

A
uniform disk with R=0.2m rolls without slipping on a horizontal surface.
String is pulled in the horizontal direction with force 15N. Moment of
inertia of disk is 0.4 kg-m2. The acceleration of the center
of the disk is most nearly

  1. 0.5 m/s2
  2. 1 m/s2
  3. 4 m/s2
  4. 7.5 m/s2
  5. 10 m/s2
  6. none of the above

Commentary:

Answer

(2) This problem can be done without knowing anything about the
friction force. To do so, though, requires knowing the Parallel Axis
Theorem for moments of inertia and the constraint between the linear and
rotational rates of motion for a rolling object. An alternate method is
to write the two equations for the linear motion of the center of mass
and the torque relation for rotation about the CM and then eliminate the
friction from the two equations.

A2L Item 188

Goal: Problem solving with rotational dynamics

Source: UMPERG-ctqpe167

A
uniform disk with mass M and radius R rolls without slipping down an
incline 30° to the horizontal. The acceleration of the center of
the disk is

  1. g/2
  2. 2g/3
  3. 3g/4
  4. g/4
  5. none of the above

Commentary:

Answer

(5) The acceleration must be smaller than for a mass sliding on a
frictionless incline, but larger than for a hoop. Application of the
rotational dynamic relation τ = Ipαp about point P, the disk’s contact
point with the incline yields an acceleration of g/3. Students must know
the moment of inertia of the disk about its center and use the Parallel
Axis Theorem.

Good discussion questions are: Would a marble have a larger or smaller
acceleration than a coin? Would the angle of the incline matter?

A2L Item 184

Goal: Problem solving with rotational dynamics

Source: UMPERG-ctqpe148

A hoop
of mass 4 kg and radius 10 cm rolls without slipping down an incline
30° to the horizontal. The acceleration of the center of the hoop
is most nearly

  1. 10 m/s2
  2. 5 m/s2
  3. 3.5 m/s2
  4. 2.5 m/s2
  5. none of the above
  6. cannot be determined

Commentary:

Answer

(4) Students should realize that the acceleration must be less
than a sliding mass on a frictionless surface would have which is #2.
Engage the students in a discussion of why the acceleration cannot
depend upon the radius.

A2L Item 178

Goal: Problem solving with rotational dynamics

Source: UMPERG-ctqpe140

A disk
on a horizontal surface sits against a curb. A string wound around the
disk is attached to a mass as shown. If R=5 cm and h=2 cm, the largest
m for which the disk will not move is

  1. Less than 2M
  2. 2M
  3. 3M
  4. 4M
  5. 5M
  6. Greater than 5M
  7. Cannot be determined.

Commentary:

Answer

(4) When m = 4M the torques about the contact point between the disk and
curb balance. Students find this problem very difficult although rather
simple. Many have the most difficulty with the simple geometry needed to
find the moment arms.

A2L Item 177

Goal: Problem solving with rotational dynamics

Source: UMPERG-ctqpe135.5

A
disk, with radius 0.25 m and mass 4 kg, lies on a smooth
horizontal table. A string wound about the disk is pulled with a
force of 8N. What is the angular acceleration of the disk about its
center?

  1. 0
  2. 64 rad/s2
  3. 8 rad/s2
  4. 4 rad/s2
  5. 12 rad/s2
  6. None of the above.
  7. Cannot be determined

Commentary:

Answer

(6) The correct value of αcm is 16 rad/s2. Students have the erroneous
concept of ‘conservation of force’. Many think that since the disk
moves, the full force cannot contribute to the torque about the center
of the disk.

A2L Item 106

Goal: Hone rotational dynamics

Source: UMPERG-ctqpe1246

A system consisting of two masses on a string is rotating with angular
velocity ω on a frictionless horizontal surface. The center of
rotation is the left-hand side of the string (nailed to the table).

The ratio of the tension in the inner string to that in the outer string
is

  1. 0.25
  2. 0.5
  3. 1.5
  4. 2.0
  5. 3.0
  6. None of the above

Commentary:

Answer

(3) Many students think the ratio is determined just by the string
lengths and give as an answer either (2) or (4). They fail to draw a
free body diagram for the inner mass and, consequently, fail to realize
that it is the net force on the inner mass that must maintain the
circular motion of the inner mass.

A2L Item 103

Goal: Reason with rotational dynamics.

Source: UMPERG-ctqpe138

A spool has string wrapped around its center axle and is sitting on a
horizontal surface. If the string is pulled at an angle to the
horizontal when drawn from the bottom of the axle, the spool will

  1. roll to the right.
  2. not roll, only slip.
  3. roll to the left.
  4. cannot be determined.

Commentary:

Answer

(4) The motion of the spool depends upon the angle θ. When the line of
action of the force passes through the contact point the spool will
slide and not rotate. At lower angles it will roll to the right and at
higher angles it will roll to the left.

A2L Item 101

Goal: Reason with rotational dynamics.

Source: UMPERG-ctqpe136

A spool has string wrapped around its center axle and is sitting on a
horizontal surface. If the string is pulled in the horizontal direction
when tangent to the top of the axle, the spool will

  1. roll to the right.
  2. not roll, only slip.
  3. roll to the left.
  4. cannot be determined.

Commentary:

Answer

(1) For many students this remains counterintuitive.

A2L Item 102

Goal: Reason with rotational dynamics.

Source: UMPERG-ctqpe137

A spool has string wrapped around its center axle and is sitting on a
horizontal surface. If the string is pulled in the horizontal direction
when tangent to the bottom of the axle, the spool will

  1. roll to the right.
  2. not roll, only slip.
  3. roll to the left.
  4. can’t be determined.

Commentary:

Answer

(1) For many students this is really counterintuitive.

A2L Item 096

Goal: Problem solving

Source: UMPERG-ctqpe166

A uniform disk with mass M and radius R sits at rest on an incline
30° to the horizontal. A string is wound around disk and attached
to top of incline as shown. The string is parallel to incline. The
friction force acting at the contact point is:

  1. Mg/2, down the incline
  2. Mg/2, up the incline
  3. Mg/4, up the incline
  4. Mg/0.86, down the incline
  5. None of the above
  6. Cannot be determined

Commentary:

Answer

(3) Balancing torques about the center of the disk determines that the
friction force points up and is equal to the tension in the string. (The
other forces, gravity and normal pass through the point and contribute
no torques.) Balancing torques about the contact point determines the
tension readily.