17. A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
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[tex]\underline \bold \red {GIVEN \: : - } \\ \\ [/tex]
[tex] \bold{ \red\mapsto \: Mass \: of \: the \: stone [m] = 0.25 kg} \\ [/tex]
[tex] \bold{\red\mapsto \: Radius [r]= 1.5m \: } \\ [/tex]
[tex] \bold{\red\mapsto \:Revolution \: 40 \: rev./min \: \: } \\ \\ [/tex]
[tex]\underline \bold \pink {TO \: FIND \: : - } \\ \\ [/tex]
[tex] \bold{ \bold \red \mapsto \: Tension \: in \: the \: string \: [Maximum \: speed \: } \\ \bold{with \: which \: the \: stone \: can \: be \: whirled} \\ \bold { around \: if \: the \: maximum \: tension \: it \: } \\ \bold{can \: withstand \: is \: 200 \: N]} \: \\ \\ [/tex]
[tex]\underline \bold \blue{FORMULA \: USED \: : - } \\ \\ [/tex]
[tex] \bold \green{ \bold \red \mapsto \: w = 2πn} \\ [/tex]
[tex] \bold \green{ \bold \red \mapsto \: T = mw ^{2}r } \\ [/tex]
[tex] \bold \red{where,} \\ [/tex]
[tex] \bold{\bold \red \mapsto \: w = Angular \: velocity.} \\
[/tex]
[tex]\bold{\bold \red \mapsto \: n= no. \: of \: revolution.} \\ [/tex]
[tex] \bold{\bold \red \mapsto \: T= Tension \: in \: the \: \: string. \: } \\ [/tex]
[tex]
\bold{\bold \red \mapsto \: m= Mass.} \: \\ \\
[/tex]
[tex]\underline \bold \green{Solution\: : - } \\ \\ [/tex]
[tex] \bold{\bold \red \mapsto \: Number \: of \: revolution \: per \: second, } \\ \bold{ \: n = \frac{40}{60} = \frac{2}{3} } \\ [/tex]
[tex] \bold{\bold \red \mapsto \: The \: centripetal \: force \: on \: stone \: is \: } \\ \bold{provided \: by \: the \: tension \: T, \: in \: } \\ \bold{the \: string,}
[/tex]
[tex] \bold \red{that \: means,} \: \\ [/tex]
[tex] \bold{ \bold \red \mapsto \: T = mw ^{2} r} \\ \\ [/tex]
[tex] \bold{Put \: values \: in \: the \: formula \: we \: get,} \\ [/tex]
[tex] \bold{ \bold \red \mapsto \: T = 0.25 \times 1.5 \times (2 \times 3.14 \times \frac{2}{3}) ^{2} } \\ [/tex]
[tex] \bold{ \bold \red \mapsto \: T = 6.57 \: N } \\ \\ [/tex]
[tex] \bold{Maximum \: tension \: in \: the \: string, } \\ \bold{Tension \: maximum \: = \: 200 N}[/tex]
[tex] \bold { \bold \red \rightarrow \:T (max) = \frac{ {mv(maz)}^{} }{r} } \\ [/tex]
[tex] \bold { \bold \red \rightarrow \:v (max) = \frac{(t(max \:)r)}{m} ^{\frac{1}{2}} } \\ [/tex]
[tex] \bold { \bold \red \rightarrow \:v (max) = ( \frac{200 \times 1.5}{0.25} )^{ \frac{1}{2} } } \\ [/tex]
[tex] \bold { \bold \red \rightarrow \:v (max) = ( \frac{300}{0.25} )^{ \frac{1}{2} } } \\ [/tex]
[tex] \bold { \bold \red \rightarrow \:v (max) = ( 1200 )^{ \frac{1}{2} } } \\ [/tex]
[tex]\bold \blue { \bold \red \rightarrow \:v (max) = 34.64 } \\ \\ [/tex]
[tex] \underline \bold \red { \bold \therefore \: max. \: speed \: of \: the \: stone \: is \: 34.64 \: m/s .\: } \\ [/tex]
Explanation:
Mass of the stone, m = 0. 25 kg
Radius of the circle r = 1. 5 m
Number of revolution per second n = 40/60 = 2/3 rps
Angular velocity
[tex]w = 2\pi \: n[/tex]
The centripetal force for the stone is provided by the tension T in the string
That's
[tex]t = m {w}^{2}r[/tex]
[tex] = 0.25 \times 1.5 \times (2 \times 3.14 \times (2|3) {)}^{2} [/tex]
[tex] = 6.57n[/tex]
Maximum tension in the string t max = 200N
[tex]tmax = \frac{ {mv}^{2max} }{t} [/tex]
[tex]v \: max = ( \frac{tmax {}^{r} }{m}) { }^{1|2} [/tex]
[tex] = = > \: (200 \times 1.5| 0.25) {}^{1|2 } [/tex]
[tex](1200) {}^{1| 2} \\ = 34.64 m| s.[/tex]
Therefore the maximum speed of the stone is 34. 64 m/s.
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