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
Menu
Mayank Kumar Jha Grade: 11
        

USING ENERGY CONSERVATION ON THE PARTICLE ONLY WE CAN THE SPEED WHEN IT IS AT THE BOTTOM POSITION.. BUT THE SOLUTION WHICH I SAW WAS A LENGTHY ONE 
 CHANGE IN PE=2*5/4R=CHANGE IN KINETIC ENERGY=1/2*MV^2 HENCE V=(5GR)^1/2

 PLZ CONFIRM..
 4. A uniform circular disc has radius R and mass m. A particle, also of mass m, is fixed t a point A on the edge of the disc as shown in the figure. The disc can rotate freely about a fixed horizontal chord PQ that is at a distance R/4 from the centre C of the disc. The line AC is perpendicular to PQ. 
 Initially the disc is held vertical with the point A at its highest position. It is the then allowed to fall so that it starts rotating about PQ. Find the linear speed of the particle as it reaches its lowest position. 

6 years ago

Answers : (1)

vikas askiitian expert
510 Points
										

solution4=>


 initially total energy (TE)  = PE + KE


 PE = PE of disk + PE of particle


taking PQ line as x axis then


PE of disk initially = mgR/4


PE of particle = mg(R+R/4) = 5mgR/4


KE initial = 0


TE =  0 + mgR/4 = 5mgR/4 = 3mgR/2       .................1


 


finally system is rotating with some angular velocity (w)


PE of disk = -mgR/4         (position of center of disk is below the x axis)


PE of particle = -5mgR/4   (position of particle is below x axis)


KE = IW2/2                  (I is moment of inertia of system about PQ)


TE = -3mgR/2 + IW2/2               ....................2


 


since total energy is conserved so equatn 1 = 2


W = [ 3mgR/I ]1/2                        ......................3


 


now I = Id (disk) + Ip(particle)


        =[mR2/4 + mR2 /16 ] (I of disk) + [ 25mR2/16 ] (I of particle)


        = 30/16mR2          


on putting this value in eq 3


W = [8g/5R]1/2


V = Wr                    (for particle , r = 5R/4)


V = [8g/5R]1/2 (5R/4) = [5gR/2]1/2


this is the required ans


approve if u like my ans

6 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: Rs. 2,756
  • View Details
  • Complete Physics Course - Class 11
  • OFFERED PRICE: Rs. 2,968
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details