Home
Class 11
PHYSICS
A circular are (AB) of thin wire frame o...

A circular are `(AB)` of thin wire frame of radius `R` and mass `M` makes an angle of `90^(@)` at the origin.The centre of mass of the arc lies at

A

`[0,(2//pi)R]`

B

`[0,(sqrt(2)//pi)R]`

C

`[0,(2sqrt(2)//pi)R]`

D

`[0,(4//pi)R]`

Text Solution

Verified by Experts

By symmetry
`x_(cm)=0`
`dm=lambdaR d theta`
`y_(cm)=(intydm)/(intdm)=(1)/(M)intRcosthetalambdaRd theta=(R^(2)lambda)/(M)int_(-pi//4)^(+pi//4)costheta d theta`
`=(R^(2)lambda)/(Rxxlambdaxx(pi//2))[2sin"(pi)/(4)]=(2R)/(pi)2(1)/(sqrt(2))=(2sqrt(2)R)/(pi)`
Promotional Banner

Topper's Solved these Questions

  • ELASTICITY & WAVES

    FIITJEE|Exercise Example|18 Videos

Similar Questions

Explore conceptually related problems

From the circular disc of radius 4R two small discs of radius R are cut off. The centre of mass of the new structure will be at

A uniform solid right circular cone of base radius R is joined to a uniform solid hemisphere of radius R and of the same density, as shown. The centre of mass of the composite solid lies at the centre of base of the cone. The height of the cone is

A piece of wire carrying a current of 6.00 A is bent in the form of a circular arc of radius 10.0 cm, and it subtends an angle of 120^@ at the centre. Find the magnetic field B due to this piece of wire at the centre.

A circular hole of radius R//2 is cut from a circular disc of radius R . The disc lies in the xy -plane and its centre coincides with the origin. If the remaining mass of the disc is M , then a. determine the initial mass of the disc and b. determine its moment of inertia about the z -axis.

The radius of gyration of a solid hemisphere of mass M and radius Rn about an axis parallel to the diameter at a distance (3)/(4) R is given by (centre of mass of the hemisphere lies at a height 3R//8 from the base.)

Four particles of masses m, 2m, 3m and 4m are arranged at the corners of a parallelogram with each side equal to a and one of the angle between two adjacent sides is 60^(@) . The parallelogram lies in the x-y plane with mass m at the origin and 4 m on the x-axis. The centre of mass of the arrangement will be located at

Mass is non - uniformly distributed on the circumference of a ring of radius a and centre at origin . Let b be the distance of centre of mass of the ring from origin. Then ,

A circular hole is cut from a disc of radius 6 cm in such a way that the radius of the hole is 1 cm and the centre of 3 cm from the centre of the disc. The distance of the centre of mass of the remaining part from the centre of the original disc is

Two stars of mass M_(1) & M_(2) are in circular orbits around their centre of mass The star of mass M_(1) has an orbit of radius R_(1) the star of mass M_(2) has an orbit of radius R_(2) (assume that their centre of mass is not acceleration and distance between starts is fixed) (a) Show that the ratio of orbital radii of the two stars equals the reciprocal of the ratio of their masses, that is R_(1)//R_(2) = M_(2)//M_(1) (b) Explain why the two stars have the same orbital period and show that the period T=2pi((R_(1)+R_(2))^(3//2))/(sqrt(G(M_(1)+M_(2)))) .

The radius of the circle whose arc of length 15\ pi cm makes an angle of (3pi)/4 radians at the centre is