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Define Gravitational Force and state its property. State and drive Newton's law of Universal Gravitation. Explain Gravitational Field and derive the expression for Gravitational Potential and Intensity.

Physics

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Step-by-step solution

6 steps

Part 1a: Moment of inertia

  1. Recall the formula for the moment of inertia of point masses rotating about an axis.

    I=m1r12+m2r22I = m_1 r_1^2 + m_2 r_2^2

  2. Convert distances to meters and substitute the given values (m1=2 kgm_1 = 2\ \mathrm{kg}, r1=10 cm=0.10 mr_1 = 10\ \mathrm{cm} = 0.10\ \mathrm{m}, m2=3 kgm_2 = 3\ \mathrm{kg}, r2=20 cm=0.20 mr_2 = 20\ \mathrm{cm} = 0.20\ \mathrm{m}).

    r12=(0.10 m)2=0.010 m2r_1^2 = (0.10\ \mathrm{m})^2 = 0.010\ \mathrm{m^2} m1r12=2 kg×0.010 m2=0.020 kg⋅m2m_1 r_1^2 = 2\ \mathrm{kg} \times 0.010\ \mathrm{m^2} = 0.020\ \mathrm{kg \cdot m^2} r22=(0.20 m)2=0.040 m2r_2^2 = (0.20\ \mathrm{m})^2 = 0.040\ \mathrm{m^2} m2r22=3 kg×0.040 m2=0.12 kg⋅m2m_2 r_2^2 = 3\ \mathrm{kg} \times 0.040\ \mathrm{m^2} = 0.12\ \mathrm{kg \cdot m^2}

  3. Add the contributions.

    I=0.020 kg⋅m2+0.12 kg⋅m2=0.14 kg⋅m2I = 0.020\ \mathrm{kg \cdot m^2} + 0.12\ \mathrm{kg \cdot m^2} = 0.14\ \mathrm{kg \cdot m^2}

    0.14 kg \cdot m^2

    Part 1b: Angular acceleration

  4. Recall the relation between torque, moment of inertia, and angular acceleration.

    τ=Iα\tau = I \alpha α=τI\alpha = \frac{\tau}{I}

  5. Substitute the given torque τ=5 N⋅m\tau = 5\ \mathrm{N \cdot m} and I=0.14 kg⋅m2I = 0.14\ \mathrm{kg \cdot m^2}.

    Express II as a fraction: 0.14=7500.14 = \frac{7}{50}.

    α=5 N⋅m750 kg⋅m2=5 N⋅m×507 kg⋅m2=2507 rad/s2\alpha = \frac{5\ \mathrm{N \cdot m}}{\frac{7}{50}\ \mathrm{kg \cdot m^2}} = 5\ \mathrm{N \cdot m} \times \frac{50}{7\ \mathrm{kg \cdot m^2}} = \frac{250}{7}\ \mathrm{rad/s^2}

  6. Approximate numerical value.

    2507≈35.7 rad/s2\frac{250}{7} \approx 35.7\ \mathrm{rad/s^2}

    \dfrac{250}{7} rad/s^2

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