{"id":1980,"date":"2018-04-20T10:19:28","date_gmt":"2018-04-20T02:19:28","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1980"},"modified":"2023-08-09T04:59:14","modified_gmt":"2023-08-09T04:59:14","slug":"rotational-motion-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/rotational-motion-problems-and-solutions.htm","title":{"rendered":"Rotational motion \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Rotational motion \u2013 problems and solutions<\/span><\/p>\n

Torque<\/b><\/span><\/p>\n

1. A beam 140 cm in length. There are three forces acts on the beam, F1<\/sub> = 20 N, F2 <\/sub>= 10 N, and F3<\/sub> = 40 N with direction and position as shown in the figure below. What is the torque<\/a> causes the beam rotates about the center of mass of the beam?<\/span><\/p>\n

Known :\"Rotational<\/u><\/span><\/p>\n

The center of mass located at the center of the beam.<\/span><\/p>\n

Length of beam (l) = 140 cm = 1.4 meters<\/span><\/p>\n

Force 1 (F1<\/sub>) = 20 N, the lever arm 1 (l1<\/sub>) = 70 cm = 0.7 meters<\/span><\/p>\n

Force 2 (F2<\/sub>) = 10 N, the lever arm 2 (l2<\/sub>) = 100 cm \u2013 70 cm = 30 cm = 0.3 meters <\/span><\/p>\n

Force 3 (F3<\/sub>) = 40 N, the lever arm 3 (l3<\/sub>) = 70 cm = 0.7 meters<\/span><\/p>\n

Wanted :<\/u> The magnitude of torque<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The torque 1 rotates beam clockwise, so assigned a negative sign to the torque 1.<\/span><\/p>\n

\u03c41<\/sub> = F1<\/sub> l1<\/sub> = (20 N)(0.7 m) = -14 N m<\/span><\/p>\n

The torque 2 rotates beam counterclockwise, so assigned a positive sign to the torque 2.<\/span><\/p>\n

\u03c42<\/sub> = F2<\/sub> l2<\/sub> = (10 N)(0.3 m) = 3 N m<\/span><\/p>\n

The torque 3 rotates beam clockwise, so assigned a positive sign to the torque 3. <\/span><\/p>\n

\u03c43<\/sub> = F3<\/sub> l3 = (40 N)(0.7 m) = -28 N m<\/span><\/p>\n

The net torque :<\/span><\/p>\n

\u03a3\u03c4 = -14 Nm + 3 Nm \u2013 28 Nm = – 42 Nm + 3 Nm = -39 Nm<\/span><\/p>\n

The magnitude of the torque is 39 N m. The direction of rotation of the beam clockwise, so assigned a negative sign.<\/span><\/p>\n

2. What is the net torque acts on the beam The axis of rotation at point D. (sin 53o<\/sup> = 0.8)<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

The axis of rotation at point D\"Rotational<\/span><\/p>\n

F1<\/sub> = 10 N and l1 <\/sub>= r1<\/sub> sin \u03b8 = (40 cm)(sin 53o<\/sup>) = (0.4 m)(0.8) = 0.32 meters<\/span><\/p>\n

F2<\/sub> = 10\u221a2 N and l2 <\/sub>= r2<\/sub> sin \u03b8 = (20 cm)(sin 45o<\/sup>) = (0.2 m)(0.5\u221a2) = 0.1\u221a2 meters<\/span><\/p>\n

F3<\/sub> = 20 N and l3<\/sub> = r1<\/sub> sin \u03b8 = (10 cm)(sin 90o<\/sup>) = (0.1 m)(1) = 0.1 meters<\/span><\/p>\n

Wanted :<\/u> The net torque<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

\u03c41<\/sub> = F1<\/sub> l1<\/sub> = (10 N)(0.32 m) = 3.2 Nm <\/span><\/p>\n

(The torque 1 rotates beam counterclockwise so we assign positive sign to the torque 1)<\/span><\/p>\n

\u03c42<\/sub> = F2<\/sub> l2<\/sub> = (10\u221a2 N)( 0.1\u221a2 m) = -2 Nm <\/span><\/p>\n

(The torque 2 rotates beam clockwise so we assign negative sign to the torque 2)<\/span><\/p>\n

\u03c43<\/sub> = F2<\/sub> l2<\/sub> = (20 N)(0.1 m) = 2 Nm <\/span><\/p>\n

(The torque 3 rotates beam counterclockwise so we assign positive sign to the torque 3)<\/span><\/p>\n

The net torque :<\/u><\/span><\/p>\n

\u03a3\u03c4 = \u03c41<\/sub> – \u03c41<\/sub> + \u03c43 <\/sub><\/span><\/p>\n

\u03a3\u03c4 = 3.2 Nm \u2013 2 Nm + 2 Nm <\/span><\/p>\n

\u03a3\u03c4 = 3.2 Nm<\/span><\/p>\n

3. What is the net torque if the axis of rotation at point D. (sin 53o <\/sup>= 0.8)<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

The axis of rotation at point D.\"Rotational<\/span><\/p>\n

Distance<\/a> between F1<\/sub> and the axis of rotation (rAD<\/sub>) = 40 cm = 0.4 m<\/span><\/p>\n

Distance between F2<\/sub> and the axis of rotation (rBD<\/sub>) = 20 cm = 0.2 m<\/span><\/p>\n

Distance between F3<\/sub> and the axis of rotation (rCD<\/sub>) = 10 cm = 0.1 m<\/span><\/p>\n

F1 <\/sub>= 10 Newton<\/span><\/p>\n

F2<\/sub> = 10\u221a2 Newton<\/span><\/p>\n

F3<\/sub> = 20 Newton<\/span><\/p>\n

Sin 53o <\/sup>= 0.8<\/span><\/p>\n

Wanted :<\/u> The net torque <\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The moment of the force <\/u>1<\/u><\/span><\/p>\n

\u03a3\u03c41<\/sub> = (F1<\/sub>)(rAD<\/sub> sin 53o<\/sup>) = (10 N)(0.4 m)(0.8) = 3.2 N.m<\/span><\/p>\n

(The torque 1 rotates beam counterclockwise so we assign positive sign to the torque 1)<\/span><\/p>\n

The moment of the force 2<\/u><\/span><\/p>\n

\u03a3\u03c42 <\/sub>= (F2<\/sub>)(rBD<\/sub> sin 45o<\/sup>) = (10\u221a2 N)(0.2 m)(0.5\u221a2) = -2 N.m<\/span><\/p>\n

(The torque 2 rotates beam clockwise so we assign negative sign to the torque 2)<\/span><\/p>\n

The moment of the force 3<\/u><\/span><\/p>\n

\u03a3\u03c43<\/sub> = (F3<\/sub>)(rCD<\/sub> sin 90o<\/sup>) = (20 N)(0.1 m)(1) = 2 N.m<\/span><\/p>\n

(The torque 2 rotates beam counterclockwise so we assign positive sign to the torque 3)<\/span><\/p>\n

The net torque :<\/u><\/span><\/p>\n

\u03a3\u03c4 = \u03a3\u03c41<\/sub> + \u03a3\u03c42 <\/sub>+ \u03a3\u03c43 <\/sub><\/span><\/p>\n

\u03a3\u03c4 = 3.2 \u2013 2 + 2<\/span><\/p>\n

\u03a3\u03c4 = 3.2 Newton meter<\/span><\/p>\n

The moment of inertia<\/b><\/span><\/p>\n

4. Length of wire = 12 m, l1<\/sub> = 4 m. Ignore wire’s mass. What is the moment of inertia<\/a> of the system.<\/span><\/p>\n

Known :\"Rotational<\/u><\/span><\/p>\n

Mass<\/a> of A (mA<\/sub>) = 0.2 kg<\/span><\/p>\n

Mass of B (mB<\/sub>) = 0.6 kg<\/span><\/p>\n

Distance between A and the axis of rotation (rA<\/sub>) = 4 meters<\/span><\/p>\n

Distance between B and the axis of rotation (rB<\/sub>) = 12 \u2013 4 = 8 meters<\/span><\/p>\n

Wanted : <\/u>The moment of inertia of the system<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The moment of inertia of A<\/u><\/span><\/p>\n

IA<\/sub> = (mA<\/sub>)(rA<\/sub>2<\/sup>) = (0.2)(4)2<\/sup> = (0.2)(16) = 3.2 kg m2<\/sup> <\/span><\/p>\n

The moment of inertia of B<\/u><\/span><\/p>\n

IB <\/sub>= (mB<\/sub>)(rB<\/sub>2<\/sup>) = (0.6)(8)2<\/sup> = (0.6)(64) = 38.4 kg m2<\/sup> <\/span><\/p>\n

The moment of inertia of the system :<\/span><\/p>\n

I = IA<\/sub> + IB<\/sub> = 3.2 + 38.4 = 41.6 kg m2<\/sup> <\/span><\/p>\n

Rotational dynamics<\/b><\/a><\/span><\/p>\n

5. A 6-N force is applied to a cord wrapped around a pulley of mass M = 5 kg and radius R = 20 cm. What is the angular acceleration of the pulley. The pulley is a uniform solid cylinder.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Force (F) = 6 Newton<\/span><\/p>\n

Mass (M) = 5 kg<\/span><\/p>\n

Radius (R) = 20 cm = 20\/100 m = 0.2 m<\/span><\/p>\n

Wanted :<\/u> Angular acceleration<\/a> (\u03b1)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The moment of the force :<\/u><\/span><\/p>\n

\u03c4 = F R = (6 Newton)(0.2 meters) = 1.2 N m<\/span><\/p>\n

The moment of inertia for solid cylinder :<\/u><\/span><\/p>\n

I = 1\/2 M R2<\/sup> <\/span><\/p>\n

I = 1\/2 (5 kg)(0.2 m)2<\/sup> <\/span><\/p>\n

I = 1\/2 (5 kg)(0.04 m2<\/sup>) <\/span><\/p>\n

I = 1\/2 (0.2) <\/span><\/p>\n

I = 0.1 kg m2<\/sup>.<\/span><\/p>\n

The angular acceleration :<\/span><\/p>\n

\u03c4 = I \u03b1<\/span><\/p>\n

\u03b1 = \u03c4 \/ I = 1.2 \/ 0.1 = 12 rad s-2 <\/sup><\/span><\/p>\n

6. A block of mass = 4 kg hanging from a cord wrapped around a pulley of mass = 8 kg and radius R = 10 cm. Acceleration due to gravity is 10 ms-2 <\/sup>. What is the linear acceleration of the block? The pulley is a uniform solid cylinder.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Mass of pulley (m) = 8 kg<\/span><\/p>\n

Radius of pulley (r) = 10 cm = 0.1 m<\/span><\/p>\n

Mass of block (m) = 4 kg<\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s2<\/sup><\/span><\/p>\n

Weight (w) = m g = (4 kg)(10 m\/s2<\/sup>) = 40 kg m\/s2<\/sup> = 40 Newton<\/span><\/p>\n

Wanted :<\/u> The free fall acceleration of the block<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The moment of inertia of the solid cylinder :<\/span><\/p>\n

I = 1\/2 M R2 <\/sup>= 1\/2 (8 kg)(0.1 m)2<\/sup> = (4 kg)(0.01 m2<\/sup>) = 0.04 kg m2 <\/sup><\/span><\/p>\n

The moment of the force :<\/span><\/p>\n

\u03c4 = F r = (40 N)(0.1 m) = 4 Nm<\/span><\/p>\n

The angular acceleration :<\/span><\/p>\n

\u03a3\u03c4 = I \u03b1 <\/span><\/p>\n

4 = 0.04 \u03b1 <\/span><\/p>\n

\u03b1 = 4 \/ 0.04 = 100<\/span><\/p>\n

The linear acceleration :<\/span><\/p>\n

a = r \u03b1 = (0.1)(100) = 10 m\/s2<\/sup><\/span><\/p>\n

7. A block with mass of m hanging from a cord wrapped around a pulley. If the free fall<\/a> acceleration of the block is a m\/s2<\/sup>, what is the moment of inertia of the pulley..<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

weight = w = m g\"Rotational<\/span><\/p>\n

Lever arm = R<\/span><\/p>\n

The angular acceleration = \u03b1 <\/span><\/p>\n

The free fall acceleration of the block = a ms-2<\/sup><\/span><\/p>\n

Wanted:<\/u> The moment of inertia of the pulley (I)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The connection between the linear acceleration and the angular acceleration :<\/span><\/p>\n

a = R \u03b1 <\/span><\/p>\n

\u03b1 = a \/ R <\/span><\/p>\n

The moment of inertia :<\/span><\/p>\n

\u03c4 = I \u03b1 <\/span><\/p>\n

I = \u03c4 : \u03b1 = \u03c4 : a \/ R = \u03c4 (R \/ a) = \u03c4 R a-1 <\/sup><\/span><\/p>\n

The angular momentum<\/b><\/span><\/p>\n

8. A 0.2-gram particle moves in a circle at a constant speed of 10 m\/s. The radius of the circle is 3 cm. What is the angular momentum of the particle?<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Mass of particle (m) = 0.2 gram = 2 x 10-4 <\/sup>kg<\/span><\/p>\n

Angular speed (\u03c9) = 10 rad s-1<\/sup><\/span><\/p>\n

Radius (r) = 3 cm = 3 x 10-2 <\/sup>meters<\/span><\/p>\n

Wanted :<\/u> The angular momentum of the particle<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The equation of the angular momentum :<\/span><\/p>\n

L = I \u03c9<\/span><\/p>\n

I = the angular momentum, I = the moment of inertia, \u03c9 = the angular speed <\/i><\/span><\/p>\n

The moment of inertia (for particle) :<\/span><\/p>\n

I = m r2 <\/sup>= (2 x 10-4 <\/sup>)(3 x 10-2<\/sup>)2 <\/sup>= (2 x 10-4 <\/sup>)(9 x 10-4<\/sup>) = 18 x 10-8<\/sup> <\/span><\/p>\n

The angular momentum :<\/span><\/p>\n

L = I \u03c9 = (18 x 10-8<\/sup>)(10 rad s-1<\/sup>) = 18 x 10-7<\/sup> kg m2<\/sup> s-1<\/sup><\/span><\/p>\n

    \n
  1. What is rotational motion?<\/strong><\/span>\n