Solved Examples on Circular & Rotational Motion

Question 1:-In Bohr’s model of hydrogen atom, an electron revolves around a proton in a circular orbit of radius 5.29×10^{-11}m with a speed of 2.18×10^{6}m/s.(a)What is the acceleration of the electron in this model of the hydrogen atom?(b)What is the magnitude and direction of the net force that acts on the electron?

Concept:

(a)The acceleration of the electron is given as:a=v^{2}/rHere,

vis the speed with which the electron revolves andris the radius of the circular orbit of electron.

(b)The force acting on the electron is the centripetal force with magnitude,F=maHere,

mis the mass of electron, and is equal to 9.1×10^{-31}kg.The force is the centripetal force, and is directed towards the center of the circular orbit.

Solution:

(a)Substitute 5.29×10^{-11}m forrand 2.18×10^{6}m/s forvin equationa=v^{2}/r,

a= (2.18×10^{6}m/s)^{2}/(5.29×10^{-11}m)= 8.98×10

^{22}m/s^{2}Therefore, the acceleration of the electron is 8.98×10

^{22}m/s^{2}.

(b)Substitute 8.98×10^{22}m/s^{2}foraand 9.1×10^{-31}kg formin equationF=ma,

Therefore, the force acting on the electron is of magnitude 8.1×10^{-8}N, and directed towards the center of the circular orbit.

Question 2:-Wheel A of radiusr_{A}= 10.0 cm is coupled by a belt b to wheelCof radiusr_{C}= 25.0 cm, as shown in below figure. WheelAincreases its angular speed from rest at a uniform rate of 1.60 rad/s^{2}. Determine the time for wheelCto 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 wheelCisωand the angular velocity of wheel_{C}Acorresponding to angular speed of wheelCisω._{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 wheelAandv_{TC}is the tangential speed of wheelC.The above equation can also be written as:

r_{A}ω_{A}=r_{C }ω_{C}

ω_{A}= (r_{C }ω_{C}) /r_{A}The time taken by wheel

Ato attain the angular speed ofω_{A}is equal to the time taken by wheelCto attain the angular speedω_{C}. Therefore if the wheel has an angular acceleration sayα, the time taken by the wheelAto move from rest toω_{A}is:

α=ω_{A}/t

t=ω_{A}/αSubstitute

ω_{A}= (r_{C }ω_{C}) /r_{A}in the equationt=ω_{A}/α,

Therefore the time taken by wheel

Cto attain the angular speed ofω_{C}ist=(r_{C }ω_{C}) /αr_{A}.

Solution:Substitute

r_{A }= 10.0 cm,r_{C }= 25.0 cm, α = 1.60 rad/s^{2}andω_{C}= 100 rev/min in equationt= (r_{C }ω_{C}) /αr_{A},

Round off to three significant figures,

t= 16.4 sTherefore the time taken by wheel

Cto attain the angular speed of 100 rev/min is 16.4 s.

Question 3:-Below figure shows a uniform block of mass,Mand edge lengthsa,b, andc. 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

Iof the block about the axis shown in figure above is1/12 [

M(a^{2}+b^{2})].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+Mh^{2}Here

his the perpendicular distance between the both the axis (refer figure below)

Solution:The distance

his 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(a^{2}+b^{2})/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.m^{2}. 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/LHere

Lis the angular momentum of the spinning top,Mis the mass of the spinning top,gis the acceleration due to gravity andris 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 equationω_{ p}=Mgr/L,

ω_{ p}=Mgr/IωHere

ωis the angular velocity of the spinning top andIits rotational inertia.

Solution:Substitute 492 g for

M, 9.81 m/s^{2}forg, 3.88 cm forr, 5.12×10^{-4}kg.m^{2}forIand 28.6 rev/s forωin equationω_{ 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.