**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}/*r*

Here, *v* is the speed with which the electron revolves and *r* is the radius of the circular orbit of electron.

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

Here, *m* is 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 for *r* and 2.18×10^{6} m/s for *v* in equation *a* = *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} for *a* and 9.1×10^{-31} kg for *m* in equation *F*= *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 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/s^{2}. 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*_{A}*ω*_{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}) /α*r*_{A}.

**Solution:**

Substitute *r*_{A }= 10.0 cm,* r*_{C }= 25.0 cm, α = 1.60 rad/s^{2} and *ω*_{C} = 100 rev/min in equation *t* * *= (*r*_{C }*ω*_{C}) /α*r*_{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* (*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 *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* (*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*/*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/s^{2} for *g*, 3.88 cm for *r*, 5.12×10^{-4} kg.m^{2} 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.