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Linear deceleration: (a_cm = -\mu_k g). Torque about CM: (\tau = f_k R = \mu_k M g R = I \alpha) with (I = \frac25 M R^2) ⇒ (\alpha = \frac5\mu_k g2R). Integrate acceleration to get velocity: v(t) = ∫(6t
Integrate acceleration to get velocity: v(t) = ∫(6t − 4) dt = 3t^2 − 4t + C. Use v(0) = −1 ⇒ C = −1. So v(t) = 3t^2 − 4t − 1 (m/s).
The force acting on the charged particle is given by:
For flat $v(r) = v_0$, enclosed mass $M(r) = \fracv_0^2 rG$. Differential rotation curve slope: $\fracdvdr = 0$ implies $\fracdMdr = \fracv_0^2G$, so $\rho(r) = \frac14\pi r^2 \fracdMdr = \fracv_0^24\pi G r^2$.