IIT JEE , AIEEE Forum - Physics , Mechanics - H.C.Verma questions..solutions needed plzzzzz..

vinayg

circular motion ques no.-17,18,19,20,21

solution needed plzzzzz..

rates will be given..

nik_gup

19) R=radius of bridge ; L = length of bridge

(a) At the highest point

mg=mv^2/r

v= Rg

(b) Given v= (1/

2)

Rg

Suppose it loses contact at B

so ,at B mgcos

= mv^2/r

v^2 =Rgcos

Rg/2 = Rgcos

cos

=1/2

=

/3

R

= R

/3

So ,it will lose contact at a distance of R/3 from highest point

(c) Let the uniform speed on the bridge be v

The chances of losing contact is max at the end of the bridge

= L/2R

now mv^2/R = mgcos

v= Rgcos(L/2R)

nik_gup

20) Since the motion is non-uniform , the acc. has both radial and

tangential components

a(r) = v^2/r

a(t) = dv/dt= a

Resultant magnitude=

(v^2/r)^2 + a^2

N = m

(v^2/r)^2 + a^2

mg = m

(v^2/r)^2 + a^2

squaring

^2 g^2 = v^4/r^2 + a^2

v^4= (

^2 g^2 - a^2)r^2

v = ^{[4 ]} (^2 g^2 - a^2)r^2

nik_gup

(a) When the ruler makes uniform circular motion in the horizontal plane

mg= m

1^2 L

1= g/L

(b) When the ruler makes uniformly accelerated circular motion

mg=

(m

2^2L)^2 + (mL

^2)^2

2^4 +

^2 =

^2 g^2/L^2

2 = ^{[4 ]}(g/L)^2 - ^2

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