Question

A particle inside a hollow sphere of radius r with coefficient of friction

Answer 1

Answer:The height is 0.134 r.

Explanation:

Given that,

Coefficient of friction \mu= \dfrac{1}{\sqrt{3}}μ=3​1​

Radius = r

According to figure,

Cos\ \theta= \dfrac{r-h}{r}Cos θ=rr−h​ .....(I)

Equivalent force is

F_{f}= \mu mg cos\thetaFf​=μmgcosθ .....(II)

F= mg sin\thetaF=mgsinθ ......(III)

Now, F_{f}=FFf​=F

\mu\ mg cos\theta = mg sin\thetaμ mgcosθ=mgsinθ

\dfrac{1}{\sqrt{3}}= tan\theta3​1​=tanθ

\theta = 30^{0}θ=300

Now, from equation (I)

cos\theta = \dfrac{r-h}{r}cosθ=rr−h​

cos 30^{0}=\dfrac{r-h}{r}cos300=rr−h​

\dfrac{\sqrt{3}}{2}= \dfrac{r-h}{r}23​​=rr−h​

h = \dfrac{(2-\sqrt{3})r}{2}h=2(2−3​)r​

h = 0.134\ rh=0.134 r

Hence, The height is 0.134 r.



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