Answered

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small block with a mass of 0.150 kg is attached to a cord passing through a hole in a frictionless, horizontal surface. The block is originally revolving a distance of 0.5 m from thehole with a seed of 0.9 m/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.23 m. a) Explain if the tension force causes the change of angular momentum. b) What is the tension in the cord in the final situation when the block moving in a circle with smaller radius. c) How much work was done by the person who pulled on the cord

Sagot :

oladayoademosu

Answer:

a. Thus, the tension in the cord causes no change in angular momentum.

b. 0.112 N

c. -0.048 J

Explanation:

a) Explain if the tension force causes the change of angular momentum.

A centripetal force F initially acts on the block to keep at at a radius of 0.5 m and speed of 0.9 m/s.

When the cord is pulled, the tension causes a change in the centripetal force to a new value at a radius of 0.23 m.

But since the force is applied in the radial direction and not perpendicular to the radius, there is no torque applied (since torque τ = rFsinθ and θ = 0 thus τ = rFsin0 = 0)and thus, there is tangential acceleration α (since α = τ/I where I = rotational inertia of block)and thus no change in angular speed ω (since α = Δω/Δt, α = 0 ⇒ Δω/Δt ⇒ Δω = 0).

Since there is a no change in angular speed, there is thus no change in angular momentum.

Thus, the tension in the cord causes no change in angular momentum.

b) What is the tension in the cord in the final situation when the block moving in a circle with smaller radius.

The centripetal force in the cord is equal to the tension at the smaller radius.

We find the angular speed of the block from v = rω where v = initial tangential speed of block at r = 0.5 m = 0.9 m/s and r = 0.5 m.

So, ω = v/r = 0.9 m/s ÷ 0.5 m = 1.8 rad/s

The centripetal force at radius r = 0.23 m is F = mrω² where m = mass of block = 0.150 kg, r = distance = 0.23 m and ω = angular speed = 1.8 rad/s

So, F = 0.150 kg × 0.23 m × (1.8 rad/s)²

F = 0.150 kg × 0.23 m × 3.24 rad²/s²

F = 0.11178 kgmrad²/s²

F ≅ 0.112 N

c) How much work was done by the person who pulled on the cord

From work-kinetic energy principles, the work done by the tension equals the kinetic energy change of the block.

ΔK = W

1/2m(v₂² - v₁²) = W where m = mass of block = 0.150 kg, v₁ = initial speed of block = 0.9 m/s, v₂ = final speed of block distance r = 0.23 m,

Since the tangential speed v ∝ r the radial distance,

v₂/v₁ = r₂/r₁

v₂ = (r₂/r₁)v₁

= 0.23 m × 0.9 m/s ÷ 0.5m

= 0.207 m²/s ÷ 0.5 m

= 0.414 m/s

So, W = 1/2m(v₂² - v₁²)

W = 1/2 × 0.150 kg((0.414 m/s)² - (0.9 m/s)²)

W = 1/2 × 0.150 kg((0.1714 m/s)² - (0.81 m/s)²)

W = 1/2 × 0.150 kg((-0.6386 m²/s²)

W = 1/2 × -0.09579 kgm²/s²)

W = -0.0479 kgm²/s²

W = -0.0479 J

W ≅ -0.048 J

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