Newton’s second law on rotational motion

4.1 The relationship between the moment of force, the moment of inertia, and the angular acceleration

If there is a resultant force (ΣF) acting on an object with mass (m) then the object moves linearly with a certain acceleration (a). The relationship between the resultant force, mass, and acceleration is expressed by the equation:

ΣF = m a

This is the equation of Newton’s second law.

The quantities of the rotational motion which are identical to the resultant force (ΣF) in linear motion is the resultant moment of force (Στ). The quantities of the rotational motion that are identical to mass (m) in linear motion is the moment of inertia (I). The quantities of the rotational motion that are identical to acceleration (a) in linear motion is the angular acceleration (α).

If there is a resultant moment of force (Στ) acting on an object that has a certain moment of inertia (I) then the object rotates with a certain angular acceleration (α). The relationship between the resultant moment force, the moment of inertia, and angular acceleration is expressed through the equation:

Στ = I α

This equation is a rotational analogy of Newton’s second law.

4.2 Sample problems of Newton’s second law on rotational motion

Sample problem 1.

Newton's second law on rotational motion 1Solid pulley with a mass of 1 kg and radius of 10 cm, on the edges, wrapped rope, one end of the rope hung with a load of 1 kg. Think the rope is massless. Determine the magnitude of the acceleration of the load when free fall downward. (g = 10 m/s2)

Solution:

Known:

F = w = m g = (1 kg)(10 m/s2) = 10 N

r = 0.1 m

Wanted: Acceleration of load

Solution:

First, calculate the moment of inertia and the moment of force.

The moment of inertia of the solid pulley:

I = 1⁄2 m r2 = 1⁄2 (1 kg)(0.1 m)2

I = (0.5 kg)(0.01 m2) = 0.005 kg m2

The moment of force:

τ = F l = (10 N)(0.1 m) = 1 N m

Angular acceleration:

Newton's second law on rotational motion 2

Acceleration of load:

a = r α = (0.1)(200) = 20 m/s2

Sample problem 2.

The solid pulley with a mass of 2M and radius of R, on the edge, wrapped a rope, one end of the rope hung with a load with a mass of m. When the load is removed, the pulley rotates with angular acceleration. If the pulley is attached to an object with mass M, so that the pulley rotates with the same angular acceleration, determine the mass of the load. (I pulley = 1⁄2 M R2).

Newton's second law on rotational motion 3Solution:

Known:

Mass of solid pulley: 2M

The radius of solid pulley: R

Mass of load: m

Wanted: Mass of load

Solution:

Calculate the moment of inertia of the solid pulley, before and after attaching objects with mass M:

The moment of inertia 1 : I = 1⁄2 m r2 = 1⁄2

(2M)(R)2 = M R2

The moment of inertia 2 : I = 1⁄2 m r2 = 1⁄2

(2M + M)(R)2 = 1⁄2 (3M)(R)2 = 1.5 M R2

Moment of the force that is exerted by the load on the pulley:

τ = F l = (m)(g)(R)

The angular acceleration of the pulley is the same, both before and after attaching objects with mass M.

Newton's second law on rotational motion 4

Mass of load = 1.5 times the mass of the original load.

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