{"id":2280,"date":"2018-04-29T14:50:22","date_gmt":"2018-04-29T06:50:22","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=2280"},"modified":"2023-08-08T13:34:37","modified_gmt":"2023-08-08T13:34:37","slug":"dynamics-of-rotational-motions-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/dynamics-of-rotational-motions-problems-and-solutions.htm","title":{"rendered":"Dynamics of rotational motions \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Dynamics of rotational motions \u2013 problems and solutions<\/span><\/p>\n

1. A pulley with the moment of inertia<\/a> I = 2\/5 MR2<\/sup> has a mass of 2-kg. If the moment of force<\/a> on the pulley is 4 N.m then what is the linear acceleration of the pulley. Acceleration due to gravity is g = 10 m.s-2<\/sup>.<\/span><\/p>\n

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

The moment of inertia of the pulley (I) = 2\/5 MR2<\/sup><\/span><\/p>\n

Mass<\/a> of pulley (M) = 2 kg<\/span><\/p>\n

Moment of force (\u03c4) = 4 Nm <\/span><\/p>\n

Acceleration due to gravity<\/a> (g) = 10 m.s-2<\/sup><\/span><\/p>\n

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

Wanted :<\/u> The linear acceleration (a)<\/span><\/p>\n

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

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

I = 2\/5 MR2 <\/sup>= 2\/5 (2)(0.2)2<\/sup> = 2\/5 (2)(0.04) = 2\/5 (0.08) = 0.16 \/ 5 = 0.032 kg m2<\/sup><\/span><\/p>\n

The angular acceleration <\/a>(\u03b1) :<\/span><\/p>\n

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

\u03b1 = \u03c4\/I = 4 \/ 0.032 = 125 rad.s-2<\/sup><\/span><\/p>\n

The linear acceleration (a) :<\/span><\/p>\n

a = R \u03b1 = (0.2)(125) = 25 m.s-2<\/sup><\/span><\/p>\n

The linear acceleration of the edge of pulley is 25 m.s-2<\/sup>. <\/span><\/p>\n

2. Acceleration of pulley is 2 ms-2<\/sup>. Acceleration due to gravity is g = 10 m.s-2<\/sup>. What is the moment of inertia of the pulley.<\/span><\/p>\n

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

The linear acceleration of the edge of pulley (a) = 2 ms-2\"Dynamics<\/sup><\/span><\/p>\n

Acceleration due to gravity (g) = 10 m.s-2<\/sup><\/span><\/p>\n

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

Object’s weight (w) = m g = (4)(10) = 40 N<\/span><\/p>\n

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

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

The moment of force (\u03c4) :<\/span><\/p>\n

\u03c4 = F R = w R = (40)(0.1) = 4 Nm<\/span><\/p>\n

The angular acceleration of the pulley (\u03b1) :<\/span><\/p>\n

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

\u03b1 = a\/R = 2 \/ 0.1 = 20 rad.s-2 <\/sup><\/span><\/p>\n

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

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

I = \u03c4\/\u03b1 = 4\/20 = 0.2 kg m2<\/sup><\/span><\/p>\n

3. The pulley accelerated at 1 ms-2<\/sup>. If radius of pulley is 10 cm,what is the moment of inertia of the pulley.<\/span><\/p>\n

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

The linear acceleration of the edge of pulley (a) = 1 ms-2\"Dynamics<\/sup><\/span><\/p>\n

Acceleration due to gravity (g) = 10 m.s-2<\/sup><\/span><\/p>\n

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

Force (F) = 4 N<\/span><\/p>\n

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

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

The moment of force (\u03c4) :<\/span><\/p>\n

\u03c4 = F R = (4)(0.1) = 0.4 Nm<\/span><\/p>\n

The angular acceleration of pulley (\u03b1) :<\/span><\/p>\n

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

\u03b1 = a\/R = 1 \/ 0.1 = 10 rad.s-2 <\/sup><\/span><\/p>\n

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

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

I = \u03c4\/\u03b1 = 0.4 \/ 10 = 0.04 kg m2<\/sup><\/span><\/p>\n

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  1. What is the rotational equivalent of mass in linear motion?<\/strong><\/span>\n