More practice questions

4. Q4. Taking the Earth’s radius as 6.37 × 10^⁶ m and one turn in 24 h, find the speed, due to the Earth’s rotation, of a point at latitude 60°.

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   ω = 2π/86 400, about 7.27 × 10^⁻⁵ rad s^⁻¹. At latitude 60° the distance from the axis (the radius of the circle of latitude) is R cos 60°, about 3.19 × 10^⁶ m, so v = ωr ≈ 7.27 × 10^⁻⁵ × 3.19 × 10^⁶, about 232 m s^⁻¹, half the speed at the equator.

5. Q5. A body rotates with ω = 3$\hat k$ rad s^⁻¹ about the z-axis. Find the velocity of the particle at r = (2$\hat i$ − $\hat j$) m, and check its magnitude with v = ωr_⊥.

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   v = 3$\hat k$ × (2$\hat i$ − $\hat j$) = 6$\hat j$ + 3$\hat i$, that is (3$\hat i$ + 6$\hat j$) m s^⁻¹, of magnitude √45, about 6.71 m s^⁻¹. Check: the particle is √5 m from the axis, and 3 × √5 is also about 6.71 m s^⁻¹.

6. Q6. A point 0.2 m from the axis of a wheel moves at 4 m s^⁻¹ and has a tangential acceleration of 2 m s^⁻². Find the magnitudes of the wheel’s angular velocity and angular acceleration, and the point’s centripetal acceleration.

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   ω = v/r = 4/0.2 = 20 rad s^⁻¹; α = a_t/r = 2/0.2 = 10 rad s^⁻²; a_c = ω^²r = 400 × 0.2 = 80 m s^⁻². The total acceleration is √(80^² + 2^²), about 80.0 m s^⁻², almost all centripetal.