Question

A particle of mass m begins to slide down a fixed smooth sphere from the top. What is its tangential acceleration when it breaks off the sphere? batado hihihi​

Answer 1

Explanation:

Suppose the particle of mass m slides down a smooth sphere of radius R, starting from rest at the top. Suppose the particle leaves the sphere at angle .

The particle slides off when the normal force is zero.

r

mv

2

=mgCosθ=−2mg(1−Cosθ)

or Cosθ=

3

2

so Sinθ=

3

5

as sin

2

θ+cos

2

θ=1

Now force in tangential direction,

F=mgSinθ=mR

Where R = a = tangential acceleration

So, a=gSinθ=g

3

5

PSP

Answer 2

Answer:

Suppose the particle of mass m slides down a smooth sphere of radius R, starting from rest at the top. Suppose the particle leaves the sphere at angle .

The particle slides off when the normal force is zero.

r

mv

2

=mgCosθ=−2mg(1−Cosθ)

or Cosθ=

3

2

so Sinθ=

3

5

as sin

2

θ+cos

2

θ=1

Now force in tangential direction,

F=mgSinθ=mR

Where R = a = tangential acceleration

So, a=gSinθ=g

3

5

Explanation:

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