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**Question 1: In Bohrâ€™s model of hydrogen atom, an electron revolves around a proton in a circular orbit of radius 5.29Ã—10 ^{-11} m with a speed of 2.18Ã—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?**

(a) The acceleration of the electron is given as: a = v

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 9.1Ã—10

The force is the centripetal force, and is directed towards the center of the circular orbit.

(a) Substitute 5.29Ã—10

a = (2.18Ã—10

= 8.98Ã—10

Therefore, the acceleration of the electron is 8.98Ã—1022 m/s2 .

(b) Substitute 8.98Ã—10

Therefore, the force acting on the electron is of magnitude 8.1Ã—10

**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.)**

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

Here v

The above equation can also be written as:

r

Ï‰

The time taken by wheel A to attain the angular speed of Ï‰A is equal to the time taken by wheel C to attain the angular speed Ï‰

Î± = Ï‰

t = Ï‰

Substitute Ï‰

Therefore the time taken by wheel C to attain the angular speed of Ï‰C is t =(r

Substitute r

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 34.0Âº with the vertical. Its mass is 492 g and its rotational inertia is 5.12Ã—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.**

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

Ï‰

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=IÏ‰. Substitute the value of angular momentum in equation Ï‰ p = Mgr/L,

Ï‰

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

Substitute 492 g for M, 9.81 m/s

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 Ï‰

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