NCERT Class 11 Physics Solutions: Chapter 7 – System of Particles and Rotational Motion Part 3

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Question 7.8:

A non-uniform bar of weight s suspended at rest by two strings of negligible weight as shown in Figure. The angles made by the strings with the vertical are and respectively. The bar is long. Calculate the distance d of the centre of gravity of the bar from its left end.

A non-uniform bar of weight W is suspended at rest by two st …

Non-Uniform Bar of Weight W Suspended by Two Strings

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Answer:

The free body diagram of the bar is shown in the following figure.

Figure shown free body diagram of the bar,where,Length of th …

Figure Shown Free Body Diagram of the Bar

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Length of the bar,

and are the tensions produced in the left and right strings respectively.

At translational equilibrium, we have:

For rotational equilibrium, on taking the torque about the centre of gravity, we have:

Hence, the . (centre of gravity) of the given bar lies from its left end.

Question 7.9:

A car weighs . The distance between its front and back axles is . Its centre of gravity is behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

Answer:

Mass of the car,

Distance between the front and back axles,

Distance between the C.G. (centre of gravity) and the back axle

The various forces acting on the car are shown in the following figure:

The various forces acting on the car are shown in the follow …

Figure Shown Various Forces Acting on the Car

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and are the forces exerted by the level ground on the front and back wheels respectively.

At translational equilibrium:

For rotational equilibrium, on taking the torque about the ., we have:

Solving equations and we get:

Therefore, the force exerted on each front wheel

The force exerted on each back wheel

Question 7.10:

(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be where is the mass of the sphere and is the radius of the sphere.

(b) Given the moment of inertia of a disc of mass M and radius about any of its diameters to be , find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Answer:

Explanation:

The moment of inertia of a sphere about its diameter

The moment of inertia (M.I.) of a sphere about its diameter

The Moment of Inertia (M.I.) of a Sphere About Its Diameter

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According to the theorem of parallel axes, the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

The M.I. about a tangent of the sphere

Explanation:

The moment of inertia of a disc about its diameter

According to the theorem of perpendicular axis, the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.

The M.I. of the disc about its centre

The situation is shown in the given figure:

Figure shown Given the moment of inertia of a disc of mass M …

Moment of Inertia of a Disc of Mass M and Radius R

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Applying the theorem of parallel axes:

The moment of inertia about an axis normal to the disc and passing through a point on

its edge