Question

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Answer 1

$\huge\underline\purple{\sf Answer:-}$

$\large{\boxed{\sf V_{o}=\sqrt{5gR}}}$

$\huge\underline\purple{\sf Solution:-}$

Given :-

Mass = m

Radius = R

To find :-

Minimum velocity with which a body of mass m enter a vertical loop = ?

━━━━━━━━━━━━━━━━━━━━━━━━━

Let ,

Tension at point A = $\sf{T_{A}}$

Using Newton's second law we known :-

$\large{\sf T_{A}-mg={\frac{mv_{c}}{R}}}$

Energy at point A = $\sf{{\frac{1}{2}}mv_{o}^{2}}$ →(1)

Energy at point C =$\sf{{\frac{1}{2}}mv_{c}^{2}+mg×2R}$ →(2)

At point C , using Newton's second law :-

$\large{\sf T_{c}+mg ={\frac{mv_{c}^{2}}{R}}}$

In order to complete a loop $\sf{T_{c}}$≥0

So ,

$\large{\sf mg={\frac{mv_{c}^{2}}{R}}}$

$\large{\sf v_{c}=\sqrt{gR}}$ →(3)

From Equation (1) and (2)

Using the Principle of Conservation Of Energy :-

$\large{\sf {\frac{1}{2}}mv_{o}^{2}={\frac{1}{2}mv_{c}^{2}}+{\frac{1}{2}}mv_{c}^{2}+2mgR}$

$\large{\sf {\frac{1}{2}}mv_{o}^{2}={\frac{1}{2}}mR+2mgR×2}$

$\large{\sf v_{o}^{2}=gR+4gR}$

$\large\red{\boxed{\sf v_{o}=\sqrt{5gR}}}$

Answer 2

Answer:

hey!

Explanation:

hey!

shrey ...

here your answer:⬇

\huge\underline\purple{\sf Answer:-}

Answer:−

\large{\boxed{\sf V_{o}=\sqrt{5gR}}}

V

o

=

5gR

\huge\underline\purple{\sf Solution:-}

Solution:−

Given :-

Mass = m

Radius = R

To find :-

Minimum velocity with which a body of mass m enter a vertical loop = ?

━━━━━━━━━━━━━━━━━━━━━━━━━

Let ,

Tension at point A = \sf{T_{A}}T

A

Using Newton's second law we known :-

\large{\sf T_{A}-mg={\frac{mv_{c}}{R}}}T

A

−mg=

R

mv

c

Energy at point A = \sf{{\frac{1}{2}}mv_{o}^{2}}

2

1

mv

o

2

→(1)

Energy at point C =\sf{{\frac{1}{2}}mv_{c}^{2}+mg×2R}

2

1

mv

c

2

+mg×2R →(2)

At point C , using Newton's second law :-

\large{\sf T_{c}+mg ={\frac{mv_{c}^{2}}{R}}}T

c

+mg=

R

mv

c

2

In order to complete a loop \sf{T_{c}}T

c

≥0

So ,

\large{\sf mg={\frac{mv_{c}^{2}}{R}}}mg=

R

mv

c

2

\large{\sf v_{c}=\sqrt{gR}}v

c

=

gR

→(3)

From Equation (1) and (2)

Using the Principle of Conservation Of Energy :-

\large{\sf {\frac{1}{2}}mv_{o}^{2}={\frac{1}{2}mv_{c}^{2}}+{\frac{1}{2}}mv_{c}^{2}+2mgR}

2

1

mv

o

2

=

2

1

mv

c

2

+

2

1

mv

c

2

+2mgR

\large{\sf {\frac{1}{2}}mv_{o}^{2}={\frac{1}{2}}mR+2mgR×2}

2

1

mv

o

2

=

2

1

mR+2mgR×2

\large{\sf v_{o}^{2}=gR+4gR}v

o

2

=gR+4gR

\large\red{\boxed{\sf v_{o}=\sqrt{5gR}}}

v

o

=

5gR