Solved Examples on Circular & Rotational Motion

Question 2:-Wheel A of radius r_A = 10.0 cm is coupled by a belt b to wheel C of radius r_C = 25.0 cm, as shown in below figure. Wheel A increases its angular speed from rest at a uniform rate of 1.60 rad/s². Determine the time for wheel C to reach a rotational speed of 100 rev/min, assuming the belt does not slip. (Hint: If the belt does not slip, the linear speeds at the rims of the two wheels must be equal.)
                                      

Concept:Assume that desired angular speed of wheel C is ω_C and the angular velocity of wheel A corresponding to angular speed of wheel C is ω_A.

The belt rolls without slipping, therefore the tangential speed of the wheel must be equal, that is v_TA = v_TC

Here v_TA is the tangential speed of wheel A and v_TC is the tangential speed of wheel C.

The above equation can also be written as:

r_Aω_A = r_C ω_C

ω_A = (r_C ω_C) /r_A

The time taken by wheel A to attain the angular speed of ω_A is equal to the time taken by wheel C to attain the angular speed ω_C. Therefore if the wheel has an angular acceleration say α, the time taken by the wheel A to move from rest to ω_A is:

 α = ω_A/t

 t = ω_A/ α

Substitute ω_A = (r_C ω_C) /r_A in the equation t = ω_A/ α,
 

Therefore the time taken by wheel C to attain the angular speed of ω_C  is t =(r_C ω_C) /αr_A.

Solution:

Substitute r_A = 10.0 cm, r_C = 25.0 cm, α = 1.60 rad/s² and ω_C = 100 rev/min in equation t   = (r_C ω_C) /αr_A,

Round off to three significant figures,

t = 16.4 s

Therefore the time taken by wheel C to attain the angular speed of 100 rev/min is 16.4 s.

Question 3:-Below figure shows a uniform block of mass, M and edge lengths a, b, and c. Calculate its rotational inertia about an axis through one corner and perpendicular to the large face of the block. 
                                                     

Concept:

Use the rotational inertia of the uniform block about the axis passing through its center and normal to the plane, as shown in the diagram above:

The rotational inertia say I of the block about the axis shown in figure above is  

1/12 [M (a²+b²)].

Therefore the rotational inertia (say I ') about the axis through one corner and perpendicular to the large face can be calculated using the parallel axis theorem as: I '= I +Mh²

Here h is the perpendicular distance between the both the axis (refer figure below) 
                                                  

Solution:

The distance h is given as:

 

Therefore the moment of inertia of the uniform block about the axis through one corner and perpendicular to large face of the block is M (a² + b²)/3.

Question 4:- A top is spinning at 28.6 rev/s about an axis making an angle of 34.0º with the vertical. Its mass is 492 g and its rotational inertia is 5.12×10^-4 kg.m². The center of mass is 3.88 cm from the pivot point. The spin is clockwise as seen from above. Find the magnitude (in rev/s) and direction of the angular velocity of precession.

Concept:

The angular velocity of the precession of the spinning top is given as:

ω _p = Mgr/L 

Here L is the angular momentum of the spinning top, M is the mass of the spinning top, g is the acceleration due to gravity and r is the distance between the origin and the point at the top.

Also, the angular momentum of the spinning top can be written as L=Iω. Substitute the value of angular momentum in equation ω _p = Mgr/L,

ω _p = Mgr / Iω

Here ω is the angular velocity of the spinning top and I its rotational inertia.

Solution:

Substitute 492 g for M, 9.81 m/s² for g, 3.88 cm for r, 5.12×10^-4 kg.m² for I and 28.6 rev/s for ω  in equation ω _p = Mgr / Iω,
 

Therefore, the angular velocity of the precession of the spinning top is 0.324 rev/s.

The figure below shows the spinning top and the direction of the angular velocity of precession.
                                                        

It can be seen from the figure that the direction of ω _p is perpendicular to the plane of cone.