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.
<u>Thus, the tension in the cord causes no change in angular momentum. </u>
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
