- #1
Karol
- 1,380
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Homework Statement
on a surface with equivalent of friction μ.
What is the friction force
The distance to rolling only
What is the energy loss in the sliding phase
Homework Equations
Kinetic energy of a solid body: ##E_k=\frac{1}{2}I\omega^2##
Torque and angular acceleration: ##M=I\alpha##
Angular velocity as a function of angular acceleration and angle: ##\omega^2=\omega_0^2+2\alpha\theta##
Angle as a function of angular acceleration and time: ##\theta=\frac{1}{2}\alpha t^2##
Shteiner's theorem: ##I_c=I_{c.o.m.}+Mr^2##
The Attempt at a Solution
$$M=I_c\alpha~~\rightarrow~~\alpha=\frac{M}{I_c}=...=\frac{g\mu}{kr}$$
$$\left\{ \begin {array} {l} \omega ^2=\omega_0^2-2\alpha \theta \\ \alpha =\frac {f\cdot r}{I_c} \\ \theta=\frac {1}{2}\alpha t^2 \end {array} \right. $$
$$\omega^2=\omega_0^2-\left( \frac{g\mu}{kr} \right)^2t^2$$
Rolling without sliding: ##\omega r=v##:
$$\left\{ \begin {array} {l} \omega^2=\omega_0^2-\left( \frac{g\mu}{kr} \right)^2t^2 \\ \omega r=v \end {array} \right. $$
$$\rightarrow \frac{v^2}{r^2}=\omega_0^2-\left( \frac{g\mu}{kr} \right)^2t^2$$
And ##v=at=g\mu\cdot t##:
$$\left\{ \begin {array} {l} \frac{v^2}{r^2}=\omega_0^2-\left( \frac{g\mu}{kr} \right)^2t^2 \\ v=g\mu\cdot t \end {array} \right. $$
$$\rightarrow ~...t^2=\frac {(v_0k)^2}{(g\mu) ^2(1+k^2)}$$
Distance:
$$s=\theta r=\frac {v_0^2k}{2g\mu (1+k^2)}$$
Shteiner's:
$$I_A=I_c+mr^2=(1+k)mr^2$$
Wasted energy:
$$\Delta E=E_i-E_f=\frac{1}{2}mv_0^2-\frac{1}{2}I_A\omega^2=...$$