A person with his hand in his pocket is skating on ice at the rate of 10m/s and describes a circleof radius 50 m. What is his inclination to vertical:(g = 10 m/sec2)a)tan–1(½)b)tan–1 (1/5)c)tan–1 (3/5)d)tan–1(1/10)Correct answer is option 'B'. Can you explain this answer?

Question

A person with his hand in his pocket is skating on ice at the rate of 10m/s and describes a circleof radius 50 m. What is his inclination to vertical:(g = 10 m/sec2)a)tan&ndash;1(&frac12;)b)tan&ndash;1 (1/5)c)tan&ndash;1 (3/5)d)tan&ndash;1(1/10)Correct answer is option 'B'. Can you explain this answer?

Answer

To find the inclination to the vertical, we need to calculate the angle between the vertical line and the line connecting the person's position to the center of the circle.

Given:
Radius of the circle, r = 50 m
Skating speed, v = 10 m/s
Acceleration due to gravity, g = 10 m/s^2

The centripetal acceleration (a_c) of the person can be calculated using the formula:
a_c = v^2 / r

Substituting the given values:
a_c = (10 m/s)^2 / 50 m
a_c = 2 m/s^2

The total acceleration experienced by the person is the sum of the centripetal acceleration (a_c) and the acceleration due to gravity (g):
a_total = a_c + g
a_total = 2 m/s^2 + 10 m/s^2
a_total = 12 m/s^2

The inclination angle (θ) can be calculated using the formula:
tan(θ) = a_total / g

Substituting the given values:
tan(θ) = 12 m/s^2 / 10 m/s^2
tan(θ) = 1.2

To find the inclination angle, we can take the inverse tangent (arctan) of 1.2:
θ = arctan(1.2)
θ ≈ 50.19 degrees

Therefore, the person's inclination to the vertical is approximately 50.19 degrees.

Answer

Since surface (ice) is frictionless, so the
centripetal force required for skating will be
provided by inclination of boy with the
vertical and that angle is given as
tan θ = v2/rg
= where v is speed of skating &
r is radius of circle in which he moves.

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