Physics, asked by aksharaa954, 8 months ago

A solid sphere of radius r rolls without slipping from rest from a height h of an inclined

track at the bottom of which there is a loop of radius R much larger than the radius of

sphere, as shown in figure. The minimum value of h for the sphere to complete the loop is​

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aksharaa954: please some one help na
aksharaa954: it's not at all clear bro

Answers

Answered by HussainSuperStudent
1

Answer:

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11th

Physics

Systems of Particles and Rotational Motion

Rolling Motion

A small solid sphere of rad...

PHYSICS

A small solid sphere of radius r rolls down an incline without slipping which ends into a vertical loop of radius R. Find the height above the base so that it just loops the loop

November 22, 2019avatar

Gurmeet Ñegī

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ANSWER

Let v

1

and w be the velocity and angular velocity of the sphere at B respectively.

As the is no friction. so conservation of energy is valid.

P.E

A

+K.E

A

=P.E

B

+K.E

B

mgh+0=mg(2R−r)+

2

1

mv

1

2

+

2

1

Iw

2

mgh=mg(2R−r)+

2

1

mv

1

2

+

2

1

×

5

2

mr

2

w

2

mgh−mg(2R−r)=

2

1

mv

1

2

+

5

1

mv

1

2

[v

1

=rw]

⟹mg(h+r−2R)=

10

7

mv

1

2

Circular motion equation:

R−r

mv

1

2

=mg

⟹mg(h+r−2R)=

10

7

mg(R−r)

⟹h=

10

27

R−

10

17

r

Explanation:

To just complete the track potential energy at height h

(Point P)=(Kinetic+ potential) energy at H

mgh=2mgR+

2

1

mv

0

2

+

2

1

0

2

where v

0

is velocity required to complete the loop

R

mv

0

2

=mg⇒v

0

2

=gR

mgh=2mgR+

2

1

mv

0

2

+

2

1

×

5

2

mr

2

×

R

v

0

2

gh=2gR+

2

1

gR+

5

1

R

2

r

2

.gR (ω

0

=

R

v

0

pure rolling)

h=

2

5R

+

5

R

[

R

2

r

2

]


aksharaa954: that's not at all clear bro
aksharaa954: need some clarrity
HussainSuperStudent: Ok Bro
HussainSuperStudent: Don't Delete
aksharaa954: yeah I din't
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