Question

A ball of mass $\sf ‘m’$ is released from a point A where, $\sf \theta_0$ is 53°. Length of pendulum is $\sf ‘l’$. Find $\sf v, a_r, a_t, T$ and $\sf F_{net}$ at
a. point A
b. point B
c. point P, where $\theta$ is 37°​

Answer 1

Q]___?

Solution-

POINT A :

=> v = 0

=> ar = v^2 / r = 0

=> at = g sinθo ...[ θo = 53° ]

= g sin 53°

= 4/5 g

anet = √[ ( ar)^2 + ( at )^2]

anet = √0t + (4/5 g )^2

anet = 4/5 g

• T = 0

• Fnet = mnet = 4/5 mg

Point B :

h = l - l cos 53°

h = l [ 1- ( 3/5) ] = 2/5 l

v^2 - u^2 = 2gh

...[ Substituting the values ]

v^2 - 0 = l × g × 2/5 l

v = √4/5 gl

=>ar = v^2 / r = (√ 4/5 gl )^2 / l = 4/5g

=>at = 0 ...[no tangential acceleration ]

=>a = √ ar^2 + at^2

=>a = √ ar^2 + 0 = ar

=>a = 4/5 g

=>T - mg = mv^2 / r

=>T - mg = [ m ( √4/5 gl )^2 ] / l

=>T = 9/5 mg

•Fnet = m × a

•Fnet = 4/5 mg

Point C :

Consider MN height down !

h = ON - OM

...[given : ON = l cos37° & OM = l cos 53°]

h = 4/5 l - 3/5 l

h = l/5

v^2- u^2 = 2gh

v^2 - 0 = 2 × g × h ...[ h = l/5]

v^2 - 0 = 2 × g × l/5

v = √2/5 gl

• ar = v^2 /r = [ (√2/5 gl )^2]/ l = 2/5 g

=> at = gsinθ = g sin 37°

=> at = 3/5 g

•a = √ ar^2 + at^2

•a = √[ (2/5 g )^2 + ( 3/5 g^2) ]

•a = √13/5 g

T - mg cos37° - mv^2 / r

T - mg ( 4/5) = m × (√2/5 gl )^2 /l

T = 6/5 mg

•Fnet = m a = m × √13/5 g

•Fnet = √13/5 mg

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Answer 2

Answer:

Q]___?

Solution-

POINT A :

=> v = 0

=> ar = v^2 / r = 0

=> at = g sinθo ...[ θo = 53° ]

= g sin 53°

= 4/5 g

anet = √[ ( ar)^2 + ( at )^2]

anet = √0t + (4/5 g )^2

anet = 4/5 g

• T = 0

• Fnet = mnet = 4/5 mg

Point B :

h = l - l cos 53°

h = l [ 1- ( 3/5) ] = 2/5 l

v^2 - u^2 = 2gh

...[ Substituting the values ]

v^2 - 0 = l × g × 2/5 l

v = √4/5 gl

=>ar = v^2 / r = (√ 4/5 gl )^2 / l = 4/5g

=>at = 0 ...[no tangential acceleration ]

=>a = √ ar^2 + at^2

=>a = √ ar^2 + 0 = ar

=>a = 4/5 g

=>T - mg = mv^2 / r

=>T - mg = [ m ( √4/5 gl )^2 ] / l

=>T = 9/5 mg

•Fnet = m × a

•Fnet = 4/5 mg

Point C :

Consider MN height down !

h = ON - OM

...[given : ON = l cos37° & OM = l cos 53°]

h = 4/5 l - 3/5 l

h = l/5

v^2- u^2 = 2gh

v^2 - 0 = 2 × g × h ...[ h = l/5]

v^2 - 0 = 2 × g × l/5

v = √2/5 gl

• ar = v^2 /r = [ (√2/5 gl )^2]/ l = 2/5 g

=> at = gsinθ = g sin 37°

=> at = 3/5 g

•a = √ ar^2 + at^2

•a = √[ (2/5 g )^2 + ( 3/5 g^2) ]

•a = √13/5 g

T - mg cos37° - mv^2 / r

T - mg ( 4/5) = m × (√2/5 gl )^2 /l

T = 6/5 mg

•Fnet = m a = m × √13/5 g

•Fnet = √13/5 mg

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