a small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. the pendulum is swinging so as to make a maximum angle of 45 ∘ with the vertical. air resistance is negligible.



Answer :

A) The rock moves at 2.1 m/s when the string passes through the vertical position. (B) 0.83 N of string tension is present when the angle is 45 degrees. (c), the string is under 1.867 N of tension as it passes through the vertical position.

We are given that,

Mass = m = 0.12kg

String length = l = 0.80m

The maximum angle = θ = 45°

a) To determine the rock's speed when the string is in the vertical position, we have

using the first principle of thermodynamics

Kinetic energy = potential energy

1/2×m×v² = m×g×l×(1-cosθ)

v² = 2×g×l×(1-cosθ)

v²  = 2×9.81×0.8×(1-cos45) = 4.597

v = √4.597 = 2.1 m/s

(b) The tension in the string at a 45° angle to the vertical is determined by in order to maintain equilibrium between tension and rock mass, ∑Forces = 0,

T - m×g×cosθ = 0

T =  m×g×cosθ

T = 0.12×9.81×cos45

T = 0.83 N

(c) The string's tension as it crosses the vertical as we cross the vertical,

T = m×g + (m×v²)/r

T = mg(1+2(1-cosθ))

T =0.981 × 0.12 (1+ 2(1-cos45))

T=1.867 N

The given question is incomplete , complete question is given below,

A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 45 ∘ with the vertical. Air resistance is negligible. (A). What is the speed of the rock when the string passes through the vertical position? Express your answer using two significant figures. (B) What is the tension in the string when it makes an angle of 45∘ with the vertical?(C). What is the tension in the string as it passes through the vertical?

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