#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# Find the force of attraction on a particle of mass m placed at the centre of a semicircular wire of length L and mass M.

Swapnil Saxena
102 Points
9 years ago

Lets consider the radius of the circle = r where r=L/pie.Consider a infinitesimally small part of the semicircular wire d (thita) located at a angle of thita form one end. The mass of this Mrd(thita)/(pie)r= Md(thita)/(pie).Then the force of attraction on the mass m due to this part of the wire is (GmM(d thita)/pie)/(r^2). Integrating the sin component of this force with respect to thita.We get (Gm M/pie)/r^2)cos(thita) from 0 to pie.= (2Gm M/pie)/r^2) putting l/(pie) in place of r. We get (2Gm M)pie/(l^2))

Arnab Mandal
18 Points
9 years ago

Can you please sent me a diagram of the sum that you have used to solve it. please!!!

Swapnil Saxena
102 Points
9 years ago

Here is the required image.