More practice questions

4. Q4. Starting from ω = ω₀ + αt, derive θ − θ₀ = ω₀t + ½αt^² by integration (NCERT leaves this as an exercise in its Example 6.10).

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   Since dθ/dt = ω₀ + αt, integrate both sides from 0 to t:

   $$\begin{aligned} \theta - \theta_0 &= \int_0^t (\omega_0 + \alpha t')\,dt' \\ &= \omega_0 t + \tfrac{1}{2}\alpha t^{2} \end{aligned}$$

5. Q5. A body starts from rest with constant angular acceleration α. Show that its angular displacement in the n-th one-second interval is α(n − ½)(1 s)^², where n is a positive integer.

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   Substitute t = n s and (n − 1) s into θ = ½αt^². Their difference is ½α[n^² − (n − 1)^²](1 s)^² = α(n − ½)(1 s)^². For α = 2 rad s^⁻² and n = 5 this gives 9 rad, as in Example 3.

6. Q6. A disc slows uniformly from 50 rad s^⁻¹ to 20 rad s^⁻¹ while turning through 35 rad. Through what further angle does it turn before stopping?

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   From ω^² = ω₀² + 2αθ: 400 = 2500 + 70α, so α = −30 rad s^⁻². From 20 rad s^⁻¹ to rest: 0 = 400 − 60θ′, so θ′ = 400/60, about 6.67 rad.