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A) i) The coefficient of friction between the block and the turn table. ii) The linear speed of the block

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Here's the solution to question 16:

16. a) i) The coefficient of friction between the block and the turn table

  1. Identify given values and convert units.

    Mass of block, m=200 g=0.2 kgm = 200 \text{ g} = 0.2 \text{ kg}
    Distance from center, r=9 cm=0.09 mr = 9 \text{ cm} = 0.09 \text{ m}
    Frictional force, Ff=1.2 NF_f = 1.2 \text{ N}
    Acceleration due to gravity, g=9.8 m/s2g = 9.8 \text{ m/s}^2

  2. Calculate the normal force (NN).

    The normal force on a horizontal surface is equal to the gravitational force.

    N=mgN = mg N=0.2 kg×9.8 m/s2N = 0.2 \text{ kg} \times 9.8 \text{ m/s}^2 N=1.96 NN = 1.96 \text{ N}

  3. Calculate the coefficient of friction (μs\mu_s).

    The maximum static frictional force is given by Ff=μsNF_f = \mu_s N.

    μs=FfN\mu_s = \frac{F_f}{N} μs=1.2 N1.96 N\mu_s = \frac{1.2 \text{ N}}{1.96 \text{ N}} μs≈0.6122\mu_s \approx 0.6122

    The coefficient of friction is 0.6122\boxed{0.6122}.

    16. a) ii) The linear speed of the block

  4. Relate frictional force to centripetal force.

    When the block begins to slide, the frictional force provides the centripetal force (FcF_c).

    Fc=Ff=1.2 NF_c = F_f = 1.2 \text{ N}

  5. Use the formula for centripetal force in terms of linear speed.

    Fc=mv2rF_c = \frac{mv^2}{r} 1.2 N=0.2 kg×v20.09 m1.2 \text{ N} = \frac{0.2 \text{ kg} \times v^2}{0.09 \text{ m}}

  6. Solve for the linear speed (vv).

    v2=1.2 N×0.09 m0.2 kgv^2 = \frac{1.2 \text{ N} \times 0.09 \text{ m}}{0.2 \text{ kg}} v2=0.1080.2v^2 = \frac{0.108}{0.2} v2=0.54 m2/s2v^2 = 0.54 \text{ m}^2/\text{s}^2 v=0.54 m2/s2v = \sqrt{0.54 \text{ m}^2/\text{s}^2} v≈0.7348 m/sv \approx 0.7348 \text{ m/s}

    The linear speed of the block is 0.7348 m/s\boxed{0.7348 \text{ m/s}}.

    16. a) iii) If the angular velocity is increased by 2π2\pi rad/s, what would be the force required to hold the block at the same place?

  7. Calculate the initial angular velocity (ω1\omega_1) when the block begins to slide.

    The centripetal force is also given by Fc=mrω2F_c = mr\omega^2.

    Fc=mrω12F_c = mr\omega_1^2 1.2 N=0.2 kg×0.09 m×ω121.2 \text{ N} = 0.2 \text{ kg} \times 0.09 \text{ m} \times \omega_1^2 1.2=0.018ω121.2 = 0.018 \omega_1^2 ω12=1.20.018\omega_1^2 = \frac{1.2}{0.018} ω12≈66.6667 rad2/s2\omega_1^2 \approx 66.6667 \text{ rad}^2/\text{s}^2 ω1=66.6667≈8.165 rad/s\omega_1 = \sqrt{66.6667} \approx 8.165 \text{ rad/s}

  8. Calculate the new angular velocity (ω2\omega_2).

    The angular velocity is increased by 2π2\pi rad/s.

    ω2=ω1+2π\omega_2 = \omega_1 + 2\pi ω2=8.165 rad/s+2×3.14159 rad/s\omega_2 = 8.165 \text{ rad/s} + 2 \times 3.14159 \text{ rad/s} ω2=8.165 rad/s+6.283 rad/s\omega_2 = 8.165 \text{ rad/s} + 6.283 \text{ rad/s} ω2≈14.448 rad/s\omega_2 \approx 14.448 \text{ rad/s}

  9. Calculate the new force required (Fc2F_{c2}) to hold the block at the same place.

    Fc2=mrω22F_{c2} = mr\omega_2^2 Fc2=0.2 kg×0.09 m×(14.448 rad/s)2F_{c2} = 0.2 \text{ kg} \times 0.09 \text{ m} \times (14.448 \text{ rad/s})^2 Fc2=0.018×208.7489F_{c2} = 0.018 \times 208.7489 Fc2≈3.757 NF_{c2} \approx 3.757 \text{ N}

    The force required to hold the block at the same place is 3.757 N\boxed{3.757 \text{ N}}.

    16. c) What is meant by the term "Banking" in roads?

    Banking in roads refers to the practice of raising the outer edge of a curved road or track relative to the inner edge. This tilt provides a component of the normal force that acts as a centripetal force, helping vehicles to negotiate turns safely at higher speeds without relying solely on friction.

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