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An aircraft with a mass of 500 kg is flying in a horizontal circle of radius 2000 m at a speed of 120 m/s. Calculate its centripetal acceleration and centripetal force. A ball with mass 4.5 x 10⁻² kg

Physics

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Answers

6 parts
d(i)

7.2 m/s2\text{7.2 m/s}^2

d(ii)

3600 N\text{3600 N}

e(i)

2.7 kg⋅m/s\text{2.7 kg}\cdot\text{m/s}

e(ii)

180 N\text{180 N}

f(ii)

10.20 m\text{10.20 m}

f(iii)

1.96 rad/s\text{1.96 rad/s}

Step-by-step solution

2 steps

Part d(i)

Step 1: Calculate the centripetal acceleration.

Given:
Mass of aircraft, m=500 kgm = 500 \text{ kg}
Radius of circular path, r=2000 mr = 2000 \text{ m}
Speed, v=120 m/sv = 120 \text{ m/s}

The formula for centripetal acceleration is ac=v2ra_c = \frac{v^2}{r}.

ac=(120 m/s)22000 ma_c = \frac{(120 \text{ m/s})^2}{2000 \text{ m}} ac=14400 m2/s22000 ma_c = \frac{14400 \text{ m}^2/\text{s}^2}{2000 \text{ m}} ac=7.2 m/s2a_c = 7.2 \text{ m/s}^2

The centripetal acceleration is 7.2 m/s2\boxed{\text{7.2 m/s}^2}.

Part d(ii)

3600 N\text{3600 N}

Step 1: Calculate the centripetal force.

The formula for centripetal force is Fc=macF_c = m a_c.

Fc=(500 kg)(7.2 m/s2)F_c = (500 \text{ kg})(7.2 \text{ m/s}^2) Fc=3600 NF_c = 3600 \text{ N}

The centripetal force is 3600 N\boxed{\text{3600 N}}.

Part e(i)

2.7 kg⋅m/s\text{2.7 kg}\cdot\text{m/s}

Step 1: Calculate the change in momentum of the ball.

Given:
Initial velocity of ball, vi=0 m/sv_i = 0 \text{ m/s} (stationary)
Final velocity of ball, vf=60 m/sv_f = 60 \text{ m/s}
Contact time, Δt=15 milliseconds=15×10−3 s\Delta t = 15 \text{ milliseconds} = 15 \times 10^{-3} \text{ s}
Mass of ball, m=4.5×10−2 kgm = 4.5 \times 10^{-2} \text{ kg}

The formula for change in momentum is Δp=m(vf−vi)\Delta p = m(v_f - v_i).

Δp=(4.5×10−2 kg)(60 m/s−0 m/s)\Delta p = (4.5 \times 10^{-2} \text{ kg})(60 \text{ m/s} - 0 \text{ m/s}) Δp=(4.5×10−2 kg)(60 m/s)\Delta p = (4.5 \times 10^{-2} \text{ kg})(60 \text{ m/s}) Δp=2.7 kg⋅m/s\Delta p = 2.7 \text{ kg}\cdot\text{m/s}

The change in momentum of the ball is 2.7 kg⋅m/s\boxed{\text{2.7 kg}\cdot\text{m/s}}.

Part e(ii)

180 N\text{180 N}

Step 1: Calculate the average force.

The formula for average force is Favg=ΔpΔtF_{avg} = \frac{\Delta p}{\Delta t}.

Favg=2.7 kg⋅m/s15×10−3 sF_{avg} = \frac{2.7 \text{ kg}\cdot\text{m/s}}{15 \times 10^{-3} \text{ s}} Favg=180 NF_{avg} = 180 \text{ N}

The average force is 180 N\boxed{\text{180 N}}.

Part f(i)

Step 1: Define centripetal force.

Given:
Weight of car, W=10,000 NW = 10,000 \text{ N}
Linear velocity, v=72 Km/hv = 72 \text{ Km/h}
Centripetal force, Fc=40,000 NF_c = 40,000 \text{ N}
Assume acceleration due to gravity, g=9.8 m/s2g = 9.8 \text{ m/s}^2.

Centripetal force is the force that acts on an object moving in a circular path, directed towards the center of the circle, and is responsible for maintaining the circular motion.

Part f(ii)

10.20 m\text{10.20 m}

Step 1: Convert linear velocity to m/s\text{m/s}.

v=72 Km/h×1000 m1 Km×1 h3600 sv = 72 \text{ Km/h} \times \frac{1000 \text{ m}}{1 \text{ Km}} \times \frac{1 \text{ h}}{3600 \text{ s}} v=20 m/sv = 20 \text{ m/s}

  1. Calculate the mass of the car.

    The formula for weight is W=mgW = mg, so m=Wgm = \frac{W}{g}.

    m=10000 N9.8 m/s2m = \frac{10000 \text{ N}}{9.8 \text{ m/s}^2} m≈1020.41 kgm \approx 1020.41 \text{ kg}

  2. Calculate the radius of the circular path.

    The formula for centripetal force is Fc=mv2rF_c = \frac{m v^2}{r}. Rearranging for rr:

    r=mv2Fcr = \frac{m v^2}{F_c} r=(1020.41 kg)(20 m/s)240000 Nr = \frac{(1020.41 \text{ kg})(20 \text{ m/s})^2}{40000 \text{ N}} r=(1020.41 kg)(400 m2/s2)40000 Nr = \frac{(1020.41 \text{ kg})(400 \text{ m}^2/\text{s}^2)}{40000 \text{ N}} r=408164 N⋅m40000 Nr = \frac{408164 \text{ N}\cdot\text{m}}{40000 \text{ N}} r≈10.20 mr \approx 10.20 \text{ m}

    The radius of the circular path is 10.20 m\boxed{\text{10.20 m}}.

Part f(iii)

1.96 rad/s\text{1.96 rad/s}

Step 1: Calculate the angular velocity of the car.

The relationship between linear velocity and angular velocity is v=rωv = r\omega. Rearranging for ω\omega:

ω=vr\omega = \frac{v}{r} ω=20 m/s10.20 m\omega = \frac{20 \text{ m/s}}{10.20 \text{ m}} ω≈1.96 rad/s\omega \approx 1.96 \text{ rad/s}

The angular velocity of the car is 1.96 rad/s\boxed{\text{1.96 rad/s}}.

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