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Newton's Laws of Motion - Limitations of Newton's Laws in Accelerating Frames-advanced

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Inertial Frame of Reference: A frame of reference that is either at rest or moving with a constant velocity. Newton's laws of motion are valid only in inertial frames.

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Non-Inertial Frame of Reference: A frame of reference that is accelerating relative to an inertial frame. In these frames, Newton's laws (specifically F=maF = ma) do not hold true in their standard form.

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The Limitation: In an accelerating (non-inertial) frame, an object appears to experience an acceleration even when no physical force is applied to it. This violates Newton's First Law.

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Pseudo Force (Fictitious Force): To apply Newton's Laws in a non-inertial frame, we must introduce an imaginary force called a Pseudo Force. It is not caused by any physical interaction.

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Direction of Pseudo Force: The pseudo force always acts in the direction opposite to the acceleration of the frame of reference (aframea_{frame}).

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Apparent Weight: In a lift (elevator) accelerating upwards or downwards, the weight measured by a scale changes because of the pseudo force acting on the body relative to the lift's frame.

📐Formulae

Fpseudo=−maframeF_{pseudo} = -m a_{frame}

Fnet=Freal+FpseudoF_{net} = F_{real} + F_{pseudo}

Wapparent=m(g+a)(Frame accelerating upwards)W_{apparent} = m(g + a) \quad \text{(Frame accelerating upwards)}

Wapparent=m(g−a)(Frame accelerating downwards)W_{apparent} = m(g - a) \quad \text{(Frame accelerating downwards)}

Wapparent=0(Free fall, where a=g)W_{apparent} = 0 \quad \text{(Free fall, where } a = g \text{)}

💡Examples

Problem 1:

A person of mass m=60 kgm = 60\text{ kg} stands on a weighing scale inside a lift that is accelerating upwards at a=3 m/s2a = 3\text{ m/s}^2. What is the reading on the weighing scale? (Take g=10 m/s2g = 10\text{ m/s}^2)

Solution:

  1. Identify the frame: The lift is accelerating upwards, so it is a non-inertial frame.
  2. Identify forces in the lift's frame:
    • True weight acting downwards: W=m×g=60×10=600 NW = m \times g = 60 \times 10 = 600\text{ N}.
    • Pseudo force acting downwards (opposite to lift's acceleration): Fp=m×a=60×3=180 NF_{p} = m \times a = 60 \times 3 = 180\text{ N}.
  3. Total downward force (Apparent Weight RR): R=W+FpR = W + F_{p} R=600+180R = 600 + 180 R=780 NR = 780\text{ N}

Explanation:

Because the lift is accelerating upwards, a pseudo force acts downwards on the person. This force adds to the actual gravitational pull, making the person feel heavier. The scale reads the Normal Force (RR), which is the apparent weight.

Problem 2:

A pendulum is hanging from the ceiling of a car. If the car accelerates forward at a=5 m/s2a = 5\text{ m/s}^2, in which direction will the pendulum bob tilt, and what is the pseudo force on a 0.2 kg0.2\text{ kg} bob?

Solution:

  1. Direction: The car is accelerating forward. In the car's frame, a pseudo force acts in the opposite direction. Therefore, the pendulum bob will tilt backwards.
  2. Magnitude of Pseudo Force: Fp=m×aF_{p} = m \times a Fp=0.2×5F_{p} = 0.2 \times 5 Fp=1.0 NF_{p} = 1.0\text{ N}

Explanation:

From the perspective of a passenger inside the car (non-inertial frame), the bob moves backward because of the pseudo force FpF_{p}. To an observer on the road (inertial frame), the bob is simply being pulled forward by the string to keep up with the car's acceleration.