Thank you for registering.

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

Click to Chat

1800-5470-145

+91 7353221155

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

509 Points
11 years ago

total change in height = dh =H-R =50cm=0.5m

from work energy theorem,

change in kinetic energy = total work

kinetic energy = mgdh

mv2(1+k2/r2)/2 = mgh

mv2= 2mgh/(1+k2/r2)    .....................1

k is radius og gyration & for sherical ball it is Rsqrt(2/5) ...

mv2 = 10mg/7           .................2

now when theta =0 then

N  = mv2/R

so N = 10mg/7R=10Newton

since the body is not falling off so net friction force should be equal to weight =mg

f = 0.7newton

nikhil arora
54 Points
11 years ago

thanks for the reply.... normal reaction is correct but frictional force is wrong by ur answer..... the right answer is 0.2 N

sorry ,i forgot to tell that it is small sphere that is is rolling down the inclined plane ....

i did it this way (but failed)

frictional force = f , alpha=angular acceletion

at  point A

f - mg = ma..... (1)

f . r = I (alpha)....(2)

putting f = I (alpha)/r in (1)

I(alpha)/r=m(g+a)

for sphere I=2mr(square)/5 and alpha=a/r

so

2mr(sqr)/5 * a/r * 1/r=m(g+a)

2/5 a= g+a

a=-5/3g

putting this in (1)

i got f=0.46

plz remove my confusion

thanks

509 Points
11 years ago

eq 1 should be

mg - f = ma ..........1

(this is because ,direction of net force on center of mass is downward ...)

f = Ia/r2

mg = 7f/2

f = 2mg/7 =0.2N ans

please approve if u like my ans