Answer to Question #306699 in Mechanics | Relativity for ED Angelo

Question #306699

A car of mass 8000kg moves in a circular path of radius 100m with a linear velocity of 50m/s.Calculate the i) Angular velocity ii) Angle the tyre makes with the ground iii) condition for skidding for u(coefficient of friction)= 0.5?

1
Expert's answer
2022-03-06T18:06:02-0500

Explanations & Calculations


a)

ω=v2r=(50ms1)2100m=25rads1\qquad\qquad \begin{aligned} \small \omega&=\small \frac{v^2}{r}\\ &=\small \frac{(50\,ms^{-1})^2}{100\,m}\\ &=\small 25\,rads^{-1} \end{aligned}


b)

Rcosθ=mv2rRsinθ=mgtanθ=rgv2θ=tan1[100×9.8502]\qquad\qquad \begin{aligned} \small R\cos\theta&=\small \frac{mv^2}{r}\\ \small R\sin\theta&=\small mg\\ \small \tan\theta&=\small \frac{rg}{v^2}\\ \small \theta&=\small \tan^{-1}\Big[\frac{100\times9.8}{50^2} \Big] \end{aligned} You can try getting the answer


c) The maximum possible speeding without skidding is when the maximum friction from the ground is possible. Apply F=ma towards the centre.

Fmax=mamaxμR=mvmax2rμmg=mvmax2rvmax=μgr=0.5×9.8×100\qquad\qquad \begin{aligned} \small F_{max}&=\small ma_{max}\\ \small \mu R&=\small m\frac{v_{max}^2}{r}\\ \small \mu mg&=\small \frac{mv_{max}^2}{r}\\ \small v_{max}&=\small \sqrt{\mu gr}\\ &=\small \sqrt{0.5\times9.8\times100} \end{aligned} Give it a try.


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