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Question-213838

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Question Number 213838 by ajfour last updated on 18/Nov/24
Commented by ajfour last updated on 18/Nov/24
A solid ball is released over a fixed cylindrical wedge as shown. Friction is sufficient. If just after the ball leaves the curved surface due to Normal reaction vanishing, it thereupon touches the ground below then find the radius ratio r/R.
$${A}\:{solid}\:{ball}\:{is}\:{released}\:{over}\:{a}\:{fixed} \\ $$$${cylindrical}\:{wedge}\:{as}\:{shown}.\:{Friction} \\ $$$${is}\:{sufficient}.\:{If}\:{just}\:{after}\:{the}\:{ball} \\ $$$${leaves}\:{the}\:{curved}\:{surface}\:{due}\:{to} \\ $$$${Normal}\:{reaction}\:{vanishing},\:{it}\: \\ $$$${thereupon}\:{touches}\:{the}\:{ground}\:{below} \\ $$$${then}\:{find}\:{the}\:{radius}\:{ratio}\:{r}/{R}. \\ $$
Answered by mr W last updated on 18/Nov/24
Commented by mr W last updated on 20/Nov/24
rotation of ball φ=(1+(R/r))θ (dφ/dt)=(1+(R/r))ω with ω=(dθ/dt) α=(d^2 φ/dt^2 )=(1+(R/r))ω×(dω/dθ) (((2mr^2 )/5)+mr^2 )×(1+(R/r))ω×(dω/dθ)=mgr sin θ ωdω=((5g)/(7(r+R))) sin θ dθ ∫_0 ^ω ωdω=((5g)/(7(r+R))) ∫_0 ^θ sin θ dθ (ω^2 /2)=((5g(1−cos θ))/(7(r+R))) ⇒ω^2 =((10g(1−cos θ))/(7(r+R))) −+−+−+−+−+−+− alternatively: mg(r+R)(1−cos θ)=(1/2)(((2mr^2 )/5))(1+(R/r))^2 ω^2 +(m/2)×r^2 ×(1+(R/r))^2 ω^2 2g(1−cos θ)=(7/5)(r+R)ω^2 ⇒ω^2 =((10g(1−cos θ))/(7(r+R))) −+−+−+−+−+−+− mg cos θ−N=(m/(r+R))×(1+(R/r))^2 r^2 ω^2 N=mg cos θ−m(r+R)×((10g(1−cos θ))/(7(r+R))) N=(((17 cos θ−10)mg)/7) N=0 ⇒17 cos θ−10=0 ⇒cos θ=((10)/(17)) ⇒θ=cos^(−1) ((10)/(17))≈53.97° θ=ϕ ⇒cos ϕ=cos θ=((10)/(17)) (r/(r+R))=((10)/(17)) ⇒(r/R)=(1/(((17)/(10))−1))=((10)/7) ✓
$${rotation}\:{of}\:{ball} \\ $$$$\phi=\left(\mathrm{1}+\frac{{R}}{{r}}\right)\theta \\ $$$$\frac{{d}\phi}{{dt}}=\left(\mathrm{1}+\frac{{R}}{{r}}\right)\omega\:{with}\:\omega=\frac{{d}\theta}{{dt}} \\ $$$$\alpha=\frac{{d}^{\mathrm{2}} \phi}{{dt}^{\mathrm{2}} }=\left(\mathrm{1}+\frac{{R}}{{r}}\right)\omega×\frac{{d}\omega}{{d}\theta} \\ $$$$\left(\frac{\mathrm{2}{mr}^{\mathrm{2}} }{\mathrm{5}}+{mr}^{\mathrm{2}} \right)×\left(\mathrm{1}+\frac{{R}}{{r}}\right)\omega×\frac{{d}\omega}{{d}\theta}={mgr}\:\mathrm{sin}\:\theta \\ $$$$\omega{d}\omega=\frac{\mathrm{5}{g}}{\mathrm{7}\left({r}+{R}\right)}\:\mathrm{sin}\:\theta\:{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\omega} \omega{d}\omega=\frac{\mathrm{5}{g}}{\mathrm{7}\left({r}+{R}\right)}\:\int_{\mathrm{0}} ^{\theta} \mathrm{sin}\:\theta\:{d}\theta \\ $$$$\frac{\omega^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{5}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{7}\left({r}+{R}\right)} \\ $$$$\Rightarrow\omega^{\mathrm{2}} =\frac{\mathrm{10}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{7}\left({r}+{R}\right)} \\ $$$$−+−+−+−+−+−+− \\ $$$${alternatively}: \\ $$$${mg}\left({r}+{R}\right)\left(\mathrm{1}−\mathrm{cos}\:\theta\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}{mr}^{\mathrm{2}} }{\mathrm{5}}\right)\left(\mathrm{1}+\frac{{R}}{{r}}\right)^{\mathrm{2}} \omega^{\mathrm{2}} +\frac{{m}}{\mathrm{2}}×{r}^{\mathrm{2}} ×\left(\mathrm{1}+\frac{{R}}{{r}}\right)^{\mathrm{2}} \omega^{\mathrm{2}} \\ $$$$\mathrm{2}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)=\frac{\mathrm{7}}{\mathrm{5}}\left({r}+{R}\right)\omega^{\mathrm{2}} \\ $$$$\Rightarrow\omega^{\mathrm{2}} =\frac{\mathrm{10}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{7}\left({r}+{R}\right)} \\ $$$$−+−+−+−+−+−+− \\ $$$${mg}\:\mathrm{cos}\:\theta−{N}=\frac{{m}}{{r}+{R}}×\left(\mathrm{1}+\frac{{R}}{{r}}\right)^{\mathrm{2}} {r}^{\mathrm{2}} \omega^{\mathrm{2}} \\ $$$${N}={mg}\:\mathrm{cos}\:\theta−{m}\left({r}+{R}\right)×\frac{\mathrm{10}{g}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{7}\left({r}+{R}\right)} \\ $$$${N}=\frac{\left(\mathrm{17}\:\mathrm{cos}\:\theta−\mathrm{10}\right){mg}}{\mathrm{7}} \\ $$$${N}=\mathrm{0}\:\Rightarrow\mathrm{17}\:\mathrm{cos}\:\theta−\mathrm{10}=\mathrm{0}\:\Rightarrow\mathrm{cos}\:\theta=\frac{\mathrm{10}}{\mathrm{17}} \\ $$$$\Rightarrow\theta=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{10}}{\mathrm{17}}\approx\mathrm{53}.\mathrm{97}° \\ $$$$\theta=\varphi\: \\ $$$$\Rightarrow\mathrm{cos}\:\varphi=\mathrm{cos}\:\theta=\frac{\mathrm{10}}{\mathrm{17}} \\ $$$$\frac{{r}}{{r}+{R}}=\frac{\mathrm{10}}{\mathrm{17}} \\ $$$$\Rightarrow\frac{{r}}{{R}}=\frac{\mathrm{1}}{\frac{\mathrm{17}}{\mathrm{10}}−\mathrm{1}}=\frac{\mathrm{10}}{\mathrm{7}}\:\checkmark \\ $$
Commented by mr W last updated on 18/Nov/24
angle θ at which the ball is losing the contact with the wedge is independent from their radii!
$${angle}\:\theta\:{at}\:{which}\:{the}\:{ball}\:{is}\:{losing} \\ $$$${the}\:{contact}\:{with}\:{the}\:{wedge}\:{is}\: \\ $$$${independent}\:{from}\:{their}\:{radii}! \\ $$
Commented by ajfour last updated on 18/Nov/24
loss in G.P.E.=gain in total K.E. mgR =(1/2)((7/5)mr^2 )ω^2 mgcos θ=mω^2 r^2 /(R+r) now dividing (R/(cos θ))=(7/(10))(R+r) now cos θ=(r/(R+r)) ⇒ (R/r)=(7/(10)) ★
$${loss}\:{in}\:{G}.{P}.{E}.={gain}\:{in}\:{total}\:{K}.{E}. \\ $$$$\:{mgR}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{7}}{\mathrm{5}}{mr}^{\mathrm{2}} \right)\omega^{\mathrm{2}} \\ $$$${mg}\mathrm{cos}\:\theta={m}\omega^{\mathrm{2}} {r}^{\mathrm{2}} /\left({R}+{r}\right) \\ $$$${now}\:{dividing} \\ $$$$\frac{{R}}{\mathrm{cos}\:\theta}=\frac{\mathrm{7}}{\mathrm{10}}\left({R}+{r}\right) \\ $$$${now}\:\:\mathrm{cos}\:\theta=\frac{{r}}{{R}+{r}} \\ $$$$\Rightarrow\:\:\frac{{R}}{{r}}=\frac{\mathrm{7}}{\mathrm{10}}\:\:\bigstar \\ $$

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