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Question Number 88623 by ajfour last updated on 11/Apr/20

Commented by ajfour last updated on 11/Apr/20

If the solid cylinder has to roll on to the inclined plane without a jump, find u_(max) .

$${If}\:{the}\:{solid}\:{cylinder}\:{has}\:{to}\:{roll} \\ $$$${on}\:{to}\:{the}\:{inclined}\:{plane}\:{without} \\ $$$${a}\:{jump},\:{find}\:{u}_{{max}} . \\ $$

Commented by mr W last updated on 13/Apr/20

it′s ok if it just happend by accident.

$${it}'{s}\:{ok}\:{if}\:{it}\:{just}\:{happend}\:{by}\:{accident}. \\ $$

Commented by Zainal Arifin last updated on 13/Apr/20

I am sorry sir

$$\mathrm{I}\:\mathrm{am}\:\mathrm{sorry}\:\mathrm{sir} \\ $$

Commented by Zainal Arifin last updated on 14/Apr/20

you are right sir. it′s by accidence.

$$\mathrm{you}\:\mathrm{are}\:\mathrm{right}\:\mathrm{sir}.\:\mathrm{it}'\mathrm{s}\:\mathrm{by}\:\mathrm{accidence}. \\ $$

Answered by mr W last updated on 12/Apr/20

Answered by ajfour last updated on 12/Apr/20

Commented by ajfour last updated on 12/Apr/20

(1/2)((3/2)mR^2 )(u_(max) ^2 /R^2 )+mgR(1−cos α) = (1/2)((3/2)mR^2 )(v^2 /R^2 ) ((mv^2 )/R)=mgcos α−N_2 for no jump and u_(max) , N_2 =0 ⇒ (3/4)mu_(max) ^2 +mgR(1−cos α) = (3/4)mgRcos α ⇒ u_(max) ^2 =(4/3){(7/4)gRcos α−gR} u_(max) =(√(((4gR)/3)((7/4)cos α−1))) . (And it is right answer, Sir!)

$$\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{3}}{\mathrm{2}}{mR}^{\mathrm{2}} \right)\frac{{u}_{{max}} ^{\mathrm{2}} }{{R}^{\mathrm{2}} }+{mgR}\left(\mathrm{1}−\mathrm{cos}\:\alpha\right) \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{3}}{\mathrm{2}}{mR}^{\mathrm{2}} \right)\frac{{v}^{\mathrm{2}} }{{R}^{\mathrm{2}} } \\ $$$$\frac{{mv}^{\mathrm{2}} }{{R}}={mg}\mathrm{cos}\:\alpha−{N}_{\mathrm{2}} \\ $$$${for}\:{no}\:{jump}\:{and}\:{u}_{{max}} \:,\:{N}_{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:\:\frac{\mathrm{3}}{\mathrm{4}}{mu}_{{max}} ^{\mathrm{2}} +{mgR}\left(\mathrm{1}−\mathrm{cos}\:\alpha\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{3}}{\mathrm{4}}{mgR}\mathrm{cos}\:\alpha \\ $$$$\Rightarrow\:\:\:{u}_{{max}} ^{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{3}}\left\{\frac{\mathrm{7}}{\mathrm{4}}{gR}\mathrm{cos}\:\alpha−{gR}\right\} \\ $$$$\:\:\:\:{u}_{{max}} =\sqrt{\frac{\mathrm{4}{gR}}{\mathrm{3}}\left(\frac{\mathrm{7}}{\mathrm{4}}\mathrm{cos}\:\alpha−\mathrm{1}\right)}\:\:. \\ $$$$\:\:\left({And}\:{it}\:{is}\:{right}\:{answer},\:{Sir}!\right) \\ $$

Commented by mr W last updated on 12/Apr/20

this is also what i thought: the cylinder should be able to rotate about the edge without losing the contact force.

$${this}\:{is}\:{also}\:{what}\:{i}\:{thought}:\:{the} \\ $$$${cylinder}\:{should}\:{be}\:{able}\:{to}\:{rotate} \\ $$$${about}\:{the}\:{edge}\:{without}\:{losing}\:{the} \\ $$$${contact}\:{force}. \\ $$

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