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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}) /αr_{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}) /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 (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.
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}) /αr_{A}.
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}) /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.
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)
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.
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.
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.
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