144031
A heavy mass is attached at one end of a thin wire and whirled in a vertical circle. The chances of breaking the wire are maximum when
1 The wire makes an angle of \(60^{\circ}\) with the horizontal
2 The mass is at the highest point of the circle
3 The mass is at the lowest point of the circle
4 The wire is horizontal
Explanation:
C The body is rotating in a vertical circle. So, at the bottom, \(\mathrm{T}_{2}=\mathrm{mg}+\mathrm{F}_{\mathrm{c}}\) At the top, \(\mathrm{T}_{1}+\mathrm{mg}=\mathrm{F}_{\mathrm{c}}\) \(\mathrm{T}_{1}=\mathrm{F}_{\mathrm{c}}-\mathrm{mg}\) Comparing equation (i) and (ii), we get - \(\mathrm{T}_{2}>\mathrm{T}_{1}\) So, tension is maximum at the lowest position of mass Hence, the chance of breaking is maximum at lowest point.
MHT-CET 2020
Motion in Plane
144032
A particle of mass ' \(m\) ' is rotating in a circle of radius ' \(r\) ' having angular momentum ' \(L\) '. Then the centripetal force will be
1 \(\frac{\mathrm{L}^{2}}{\mathrm{mr}^{2}}\)
2 \(\frac{\mathrm{L}^{2}}{\mathrm{mr}^{2}}\)
3 \(\frac{\mathrm{L}^{2}}{\mathrm{mr}}\)
4 \(\frac{\mathrm{L}^{2}}{\mathrm{mr}^{3}}\)
Explanation:
D Given, mass \(=\mathrm{m}\), circle radius \(=\mathrm{r}\), angular momentum \(=\mathrm{L}\) We know, Centripetal force \(\left(\mathrm{F}_{\mathrm{c}}\right)=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) Angular momentum \((\mathrm{L})=\mathrm{mvr}\) \(\mathrm{v}=\frac{\mathrm{L}}{\mathrm{mr}}\) Substituting the value of ' \(v\) ' from equation (ii) to equation (i), \(\mathrm{F}_{\mathrm{c}}=\frac{\mathrm{m}}{\mathrm{r}}\left(\frac{\mathrm{L}}{\mathrm{mr}}\right)^{2}\) \(\mathrm{~F}_{\mathrm{c}}=\frac{\mathrm{m}}{\mathrm{r}} \times \frac{\mathrm{L}^{2}}{\mathrm{~m}^{2} \mathrm{r}^{2}} \Rightarrow \mathrm{F}_{\mathrm{c}}=\frac{\mathrm{L}^{2}}{\mathrm{mr}^{3}}\)
MHT-CET 2020
Motion in Plane
144034
A particle is moving in a circle of radius ' \(R\) ' with constant speed ' \(V\) '. The magnitude of average acceleration after half revolution is
1 \(\frac{2 V^{2}}{\pi R}\)
2 \(\frac{2 \pi}{R V^{2}}\)
3 \(\frac{2 \mathrm{~V}}{\pi \mathrm{R}^{2}}\)
4 \(\frac{2 \mathrm{R}}{\pi \mathrm{V}}\)
Explanation:
A In a half revolution, change in momentum, \(\Delta \mathrm{P}=\mathrm{mV}-\mathrm{m}(-\mathrm{V})=2 \mathrm{mV}\) The amount of time it takes for a particle to complete a half revolution, \(\mathrm{t}=\frac{\pi \mathrm{R}}{\mathrm{V}}\) Average force \((F)=\) Rate of change in momentum \(=\frac{\Delta \mathrm{P}}{\mathrm{t}}\) \(=\frac{2 \mathrm{mV}}{\frac{\pi \mathrm{R}}{\mathrm{V}}} \quad \text { [From equation (i) and (ii)] }\) \(\mathrm{F}=\frac{2 m \mathrm{~V}^{2}}{\pi \mathrm{R}} \quad\{\because \mathrm{F}=\mathrm{ma}\}\) \(\mathrm{ma}=\frac{2 \mathrm{mV} \mathrm{V}^{2}}{\pi \mathrm{R}} \quad\) \(\mathrm{a}=\frac{2 \mathrm{~V}^{2}}{\pi \mathrm{R}}\)
MHT-CET 2020
Motion in Plane
144035
A particle of mass ' \(m\) ' is performing U.C.M. along a circle of radius ' \(r\) '. The relation between centripetal acceleration ' \(a\) ' and kinetic energy ' \(E\) ' is given by
C We know that \(\mathrm{K} . \mathrm{E}=\frac{1}{2} \mathrm{I} \omega^{2}\) and \(\quad \omega^{2}=\frac{\mathrm{a}}{\mathrm{r}}\) from eqn (i) and (ii) \(\therefore \quad \mathrm{E}=\frac{\mathrm{I}}{2} \times \frac{\mathrm{a}}{\mathrm{r}}\) \(\mathrm{E}=\frac{1}{2} \mathrm{mr}^{2} \times \frac{\mathrm{a}}{\mathrm{r}} \quad\left[\mathrm{I}=\mathrm{mr}^{2}\right]\) \(\mathrm{E}=\frac{1}{2} \mathrm{mr} \cdot \mathrm{a}\) \(\mathrm{a}=\frac{2 \mathrm{E}}{\mathrm{mr}}\)
MHT-CET 2020
Motion in Plane
144036
A train has to negotiate a curve of radius ' \(r\) ' \(m\), the distance between the rails is ' \(\ell\) ' \(\mathrm{m}\) and outer rail is raised above inner rail by distance of ' \(h\) ' \(m\). If the angle of banking is small, the safety speed limit on this banked road is
144031
A heavy mass is attached at one end of a thin wire and whirled in a vertical circle. The chances of breaking the wire are maximum when
1 The wire makes an angle of \(60^{\circ}\) with the horizontal
2 The mass is at the highest point of the circle
3 The mass is at the lowest point of the circle
4 The wire is horizontal
Explanation:
C The body is rotating in a vertical circle. So, at the bottom, \(\mathrm{T}_{2}=\mathrm{mg}+\mathrm{F}_{\mathrm{c}}\) At the top, \(\mathrm{T}_{1}+\mathrm{mg}=\mathrm{F}_{\mathrm{c}}\) \(\mathrm{T}_{1}=\mathrm{F}_{\mathrm{c}}-\mathrm{mg}\) Comparing equation (i) and (ii), we get - \(\mathrm{T}_{2}>\mathrm{T}_{1}\) So, tension is maximum at the lowest position of mass Hence, the chance of breaking is maximum at lowest point.
MHT-CET 2020
Motion in Plane
144032
A particle of mass ' \(m\) ' is rotating in a circle of radius ' \(r\) ' having angular momentum ' \(L\) '. Then the centripetal force will be
1 \(\frac{\mathrm{L}^{2}}{\mathrm{mr}^{2}}\)
2 \(\frac{\mathrm{L}^{2}}{\mathrm{mr}^{2}}\)
3 \(\frac{\mathrm{L}^{2}}{\mathrm{mr}}\)
4 \(\frac{\mathrm{L}^{2}}{\mathrm{mr}^{3}}\)
Explanation:
D Given, mass \(=\mathrm{m}\), circle radius \(=\mathrm{r}\), angular momentum \(=\mathrm{L}\) We know, Centripetal force \(\left(\mathrm{F}_{\mathrm{c}}\right)=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) Angular momentum \((\mathrm{L})=\mathrm{mvr}\) \(\mathrm{v}=\frac{\mathrm{L}}{\mathrm{mr}}\) Substituting the value of ' \(v\) ' from equation (ii) to equation (i), \(\mathrm{F}_{\mathrm{c}}=\frac{\mathrm{m}}{\mathrm{r}}\left(\frac{\mathrm{L}}{\mathrm{mr}}\right)^{2}\) \(\mathrm{~F}_{\mathrm{c}}=\frac{\mathrm{m}}{\mathrm{r}} \times \frac{\mathrm{L}^{2}}{\mathrm{~m}^{2} \mathrm{r}^{2}} \Rightarrow \mathrm{F}_{\mathrm{c}}=\frac{\mathrm{L}^{2}}{\mathrm{mr}^{3}}\)
MHT-CET 2020
Motion in Plane
144034
A particle is moving in a circle of radius ' \(R\) ' with constant speed ' \(V\) '. The magnitude of average acceleration after half revolution is
1 \(\frac{2 V^{2}}{\pi R}\)
2 \(\frac{2 \pi}{R V^{2}}\)
3 \(\frac{2 \mathrm{~V}}{\pi \mathrm{R}^{2}}\)
4 \(\frac{2 \mathrm{R}}{\pi \mathrm{V}}\)
Explanation:
A In a half revolution, change in momentum, \(\Delta \mathrm{P}=\mathrm{mV}-\mathrm{m}(-\mathrm{V})=2 \mathrm{mV}\) The amount of time it takes for a particle to complete a half revolution, \(\mathrm{t}=\frac{\pi \mathrm{R}}{\mathrm{V}}\) Average force \((F)=\) Rate of change in momentum \(=\frac{\Delta \mathrm{P}}{\mathrm{t}}\) \(=\frac{2 \mathrm{mV}}{\frac{\pi \mathrm{R}}{\mathrm{V}}} \quad \text { [From equation (i) and (ii)] }\) \(\mathrm{F}=\frac{2 m \mathrm{~V}^{2}}{\pi \mathrm{R}} \quad\{\because \mathrm{F}=\mathrm{ma}\}\) \(\mathrm{ma}=\frac{2 \mathrm{mV} \mathrm{V}^{2}}{\pi \mathrm{R}} \quad\) \(\mathrm{a}=\frac{2 \mathrm{~V}^{2}}{\pi \mathrm{R}}\)
MHT-CET 2020
Motion in Plane
144035
A particle of mass ' \(m\) ' is performing U.C.M. along a circle of radius ' \(r\) '. The relation between centripetal acceleration ' \(a\) ' and kinetic energy ' \(E\) ' is given by
C We know that \(\mathrm{K} . \mathrm{E}=\frac{1}{2} \mathrm{I} \omega^{2}\) and \(\quad \omega^{2}=\frac{\mathrm{a}}{\mathrm{r}}\) from eqn (i) and (ii) \(\therefore \quad \mathrm{E}=\frac{\mathrm{I}}{2} \times \frac{\mathrm{a}}{\mathrm{r}}\) \(\mathrm{E}=\frac{1}{2} \mathrm{mr}^{2} \times \frac{\mathrm{a}}{\mathrm{r}} \quad\left[\mathrm{I}=\mathrm{mr}^{2}\right]\) \(\mathrm{E}=\frac{1}{2} \mathrm{mr} \cdot \mathrm{a}\) \(\mathrm{a}=\frac{2 \mathrm{E}}{\mathrm{mr}}\)
MHT-CET 2020
Motion in Plane
144036
A train has to negotiate a curve of radius ' \(r\) ' \(m\), the distance between the rails is ' \(\ell\) ' \(\mathrm{m}\) and outer rail is raised above inner rail by distance of ' \(h\) ' \(m\). If the angle of banking is small, the safety speed limit on this banked road is
144031
A heavy mass is attached at one end of a thin wire and whirled in a vertical circle. The chances of breaking the wire are maximum when
1 The wire makes an angle of \(60^{\circ}\) with the horizontal
2 The mass is at the highest point of the circle
3 The mass is at the lowest point of the circle
4 The wire is horizontal
Explanation:
C The body is rotating in a vertical circle. So, at the bottom, \(\mathrm{T}_{2}=\mathrm{mg}+\mathrm{F}_{\mathrm{c}}\) At the top, \(\mathrm{T}_{1}+\mathrm{mg}=\mathrm{F}_{\mathrm{c}}\) \(\mathrm{T}_{1}=\mathrm{F}_{\mathrm{c}}-\mathrm{mg}\) Comparing equation (i) and (ii), we get - \(\mathrm{T}_{2}>\mathrm{T}_{1}\) So, tension is maximum at the lowest position of mass Hence, the chance of breaking is maximum at lowest point.
MHT-CET 2020
Motion in Plane
144032
A particle of mass ' \(m\) ' is rotating in a circle of radius ' \(r\) ' having angular momentum ' \(L\) '. Then the centripetal force will be
1 \(\frac{\mathrm{L}^{2}}{\mathrm{mr}^{2}}\)
2 \(\frac{\mathrm{L}^{2}}{\mathrm{mr}^{2}}\)
3 \(\frac{\mathrm{L}^{2}}{\mathrm{mr}}\)
4 \(\frac{\mathrm{L}^{2}}{\mathrm{mr}^{3}}\)
Explanation:
D Given, mass \(=\mathrm{m}\), circle radius \(=\mathrm{r}\), angular momentum \(=\mathrm{L}\) We know, Centripetal force \(\left(\mathrm{F}_{\mathrm{c}}\right)=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) Angular momentum \((\mathrm{L})=\mathrm{mvr}\) \(\mathrm{v}=\frac{\mathrm{L}}{\mathrm{mr}}\) Substituting the value of ' \(v\) ' from equation (ii) to equation (i), \(\mathrm{F}_{\mathrm{c}}=\frac{\mathrm{m}}{\mathrm{r}}\left(\frac{\mathrm{L}}{\mathrm{mr}}\right)^{2}\) \(\mathrm{~F}_{\mathrm{c}}=\frac{\mathrm{m}}{\mathrm{r}} \times \frac{\mathrm{L}^{2}}{\mathrm{~m}^{2} \mathrm{r}^{2}} \Rightarrow \mathrm{F}_{\mathrm{c}}=\frac{\mathrm{L}^{2}}{\mathrm{mr}^{3}}\)
MHT-CET 2020
Motion in Plane
144034
A particle is moving in a circle of radius ' \(R\) ' with constant speed ' \(V\) '. The magnitude of average acceleration after half revolution is
1 \(\frac{2 V^{2}}{\pi R}\)
2 \(\frac{2 \pi}{R V^{2}}\)
3 \(\frac{2 \mathrm{~V}}{\pi \mathrm{R}^{2}}\)
4 \(\frac{2 \mathrm{R}}{\pi \mathrm{V}}\)
Explanation:
A In a half revolution, change in momentum, \(\Delta \mathrm{P}=\mathrm{mV}-\mathrm{m}(-\mathrm{V})=2 \mathrm{mV}\) The amount of time it takes for a particle to complete a half revolution, \(\mathrm{t}=\frac{\pi \mathrm{R}}{\mathrm{V}}\) Average force \((F)=\) Rate of change in momentum \(=\frac{\Delta \mathrm{P}}{\mathrm{t}}\) \(=\frac{2 \mathrm{mV}}{\frac{\pi \mathrm{R}}{\mathrm{V}}} \quad \text { [From equation (i) and (ii)] }\) \(\mathrm{F}=\frac{2 m \mathrm{~V}^{2}}{\pi \mathrm{R}} \quad\{\because \mathrm{F}=\mathrm{ma}\}\) \(\mathrm{ma}=\frac{2 \mathrm{mV} \mathrm{V}^{2}}{\pi \mathrm{R}} \quad\) \(\mathrm{a}=\frac{2 \mathrm{~V}^{2}}{\pi \mathrm{R}}\)
MHT-CET 2020
Motion in Plane
144035
A particle of mass ' \(m\) ' is performing U.C.M. along a circle of radius ' \(r\) '. The relation between centripetal acceleration ' \(a\) ' and kinetic energy ' \(E\) ' is given by
C We know that \(\mathrm{K} . \mathrm{E}=\frac{1}{2} \mathrm{I} \omega^{2}\) and \(\quad \omega^{2}=\frac{\mathrm{a}}{\mathrm{r}}\) from eqn (i) and (ii) \(\therefore \quad \mathrm{E}=\frac{\mathrm{I}}{2} \times \frac{\mathrm{a}}{\mathrm{r}}\) \(\mathrm{E}=\frac{1}{2} \mathrm{mr}^{2} \times \frac{\mathrm{a}}{\mathrm{r}} \quad\left[\mathrm{I}=\mathrm{mr}^{2}\right]\) \(\mathrm{E}=\frac{1}{2} \mathrm{mr} \cdot \mathrm{a}\) \(\mathrm{a}=\frac{2 \mathrm{E}}{\mathrm{mr}}\)
MHT-CET 2020
Motion in Plane
144036
A train has to negotiate a curve of radius ' \(r\) ' \(m\), the distance between the rails is ' \(\ell\) ' \(\mathrm{m}\) and outer rail is raised above inner rail by distance of ' \(h\) ' \(m\). If the angle of banking is small, the safety speed limit on this banked road is
144031
A heavy mass is attached at one end of a thin wire and whirled in a vertical circle. The chances of breaking the wire are maximum when
1 The wire makes an angle of \(60^{\circ}\) with the horizontal
2 The mass is at the highest point of the circle
3 The mass is at the lowest point of the circle
4 The wire is horizontal
Explanation:
C The body is rotating in a vertical circle. So, at the bottom, \(\mathrm{T}_{2}=\mathrm{mg}+\mathrm{F}_{\mathrm{c}}\) At the top, \(\mathrm{T}_{1}+\mathrm{mg}=\mathrm{F}_{\mathrm{c}}\) \(\mathrm{T}_{1}=\mathrm{F}_{\mathrm{c}}-\mathrm{mg}\) Comparing equation (i) and (ii), we get - \(\mathrm{T}_{2}>\mathrm{T}_{1}\) So, tension is maximum at the lowest position of mass Hence, the chance of breaking is maximum at lowest point.
MHT-CET 2020
Motion in Plane
144032
A particle of mass ' \(m\) ' is rotating in a circle of radius ' \(r\) ' having angular momentum ' \(L\) '. Then the centripetal force will be
1 \(\frac{\mathrm{L}^{2}}{\mathrm{mr}^{2}}\)
2 \(\frac{\mathrm{L}^{2}}{\mathrm{mr}^{2}}\)
3 \(\frac{\mathrm{L}^{2}}{\mathrm{mr}}\)
4 \(\frac{\mathrm{L}^{2}}{\mathrm{mr}^{3}}\)
Explanation:
D Given, mass \(=\mathrm{m}\), circle radius \(=\mathrm{r}\), angular momentum \(=\mathrm{L}\) We know, Centripetal force \(\left(\mathrm{F}_{\mathrm{c}}\right)=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) Angular momentum \((\mathrm{L})=\mathrm{mvr}\) \(\mathrm{v}=\frac{\mathrm{L}}{\mathrm{mr}}\) Substituting the value of ' \(v\) ' from equation (ii) to equation (i), \(\mathrm{F}_{\mathrm{c}}=\frac{\mathrm{m}}{\mathrm{r}}\left(\frac{\mathrm{L}}{\mathrm{mr}}\right)^{2}\) \(\mathrm{~F}_{\mathrm{c}}=\frac{\mathrm{m}}{\mathrm{r}} \times \frac{\mathrm{L}^{2}}{\mathrm{~m}^{2} \mathrm{r}^{2}} \Rightarrow \mathrm{F}_{\mathrm{c}}=\frac{\mathrm{L}^{2}}{\mathrm{mr}^{3}}\)
MHT-CET 2020
Motion in Plane
144034
A particle is moving in a circle of radius ' \(R\) ' with constant speed ' \(V\) '. The magnitude of average acceleration after half revolution is
1 \(\frac{2 V^{2}}{\pi R}\)
2 \(\frac{2 \pi}{R V^{2}}\)
3 \(\frac{2 \mathrm{~V}}{\pi \mathrm{R}^{2}}\)
4 \(\frac{2 \mathrm{R}}{\pi \mathrm{V}}\)
Explanation:
A In a half revolution, change in momentum, \(\Delta \mathrm{P}=\mathrm{mV}-\mathrm{m}(-\mathrm{V})=2 \mathrm{mV}\) The amount of time it takes for a particle to complete a half revolution, \(\mathrm{t}=\frac{\pi \mathrm{R}}{\mathrm{V}}\) Average force \((F)=\) Rate of change in momentum \(=\frac{\Delta \mathrm{P}}{\mathrm{t}}\) \(=\frac{2 \mathrm{mV}}{\frac{\pi \mathrm{R}}{\mathrm{V}}} \quad \text { [From equation (i) and (ii)] }\) \(\mathrm{F}=\frac{2 m \mathrm{~V}^{2}}{\pi \mathrm{R}} \quad\{\because \mathrm{F}=\mathrm{ma}\}\) \(\mathrm{ma}=\frac{2 \mathrm{mV} \mathrm{V}^{2}}{\pi \mathrm{R}} \quad\) \(\mathrm{a}=\frac{2 \mathrm{~V}^{2}}{\pi \mathrm{R}}\)
MHT-CET 2020
Motion in Plane
144035
A particle of mass ' \(m\) ' is performing U.C.M. along a circle of radius ' \(r\) '. The relation between centripetal acceleration ' \(a\) ' and kinetic energy ' \(E\) ' is given by
C We know that \(\mathrm{K} . \mathrm{E}=\frac{1}{2} \mathrm{I} \omega^{2}\) and \(\quad \omega^{2}=\frac{\mathrm{a}}{\mathrm{r}}\) from eqn (i) and (ii) \(\therefore \quad \mathrm{E}=\frac{\mathrm{I}}{2} \times \frac{\mathrm{a}}{\mathrm{r}}\) \(\mathrm{E}=\frac{1}{2} \mathrm{mr}^{2} \times \frac{\mathrm{a}}{\mathrm{r}} \quad\left[\mathrm{I}=\mathrm{mr}^{2}\right]\) \(\mathrm{E}=\frac{1}{2} \mathrm{mr} \cdot \mathrm{a}\) \(\mathrm{a}=\frac{2 \mathrm{E}}{\mathrm{mr}}\)
MHT-CET 2020
Motion in Plane
144036
A train has to negotiate a curve of radius ' \(r\) ' \(m\), the distance between the rails is ' \(\ell\) ' \(\mathrm{m}\) and outer rail is raised above inner rail by distance of ' \(h\) ' \(m\). If the angle of banking is small, the safety speed limit on this banked road is
144031
A heavy mass is attached at one end of a thin wire and whirled in a vertical circle. The chances of breaking the wire are maximum when
1 The wire makes an angle of \(60^{\circ}\) with the horizontal
2 The mass is at the highest point of the circle
3 The mass is at the lowest point of the circle
4 The wire is horizontal
Explanation:
C The body is rotating in a vertical circle. So, at the bottom, \(\mathrm{T}_{2}=\mathrm{mg}+\mathrm{F}_{\mathrm{c}}\) At the top, \(\mathrm{T}_{1}+\mathrm{mg}=\mathrm{F}_{\mathrm{c}}\) \(\mathrm{T}_{1}=\mathrm{F}_{\mathrm{c}}-\mathrm{mg}\) Comparing equation (i) and (ii), we get - \(\mathrm{T}_{2}>\mathrm{T}_{1}\) So, tension is maximum at the lowest position of mass Hence, the chance of breaking is maximum at lowest point.
MHT-CET 2020
Motion in Plane
144032
A particle of mass ' \(m\) ' is rotating in a circle of radius ' \(r\) ' having angular momentum ' \(L\) '. Then the centripetal force will be
1 \(\frac{\mathrm{L}^{2}}{\mathrm{mr}^{2}}\)
2 \(\frac{\mathrm{L}^{2}}{\mathrm{mr}^{2}}\)
3 \(\frac{\mathrm{L}^{2}}{\mathrm{mr}}\)
4 \(\frac{\mathrm{L}^{2}}{\mathrm{mr}^{3}}\)
Explanation:
D Given, mass \(=\mathrm{m}\), circle radius \(=\mathrm{r}\), angular momentum \(=\mathrm{L}\) We know, Centripetal force \(\left(\mathrm{F}_{\mathrm{c}}\right)=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) Angular momentum \((\mathrm{L})=\mathrm{mvr}\) \(\mathrm{v}=\frac{\mathrm{L}}{\mathrm{mr}}\) Substituting the value of ' \(v\) ' from equation (ii) to equation (i), \(\mathrm{F}_{\mathrm{c}}=\frac{\mathrm{m}}{\mathrm{r}}\left(\frac{\mathrm{L}}{\mathrm{mr}}\right)^{2}\) \(\mathrm{~F}_{\mathrm{c}}=\frac{\mathrm{m}}{\mathrm{r}} \times \frac{\mathrm{L}^{2}}{\mathrm{~m}^{2} \mathrm{r}^{2}} \Rightarrow \mathrm{F}_{\mathrm{c}}=\frac{\mathrm{L}^{2}}{\mathrm{mr}^{3}}\)
MHT-CET 2020
Motion in Plane
144034
A particle is moving in a circle of radius ' \(R\) ' with constant speed ' \(V\) '. The magnitude of average acceleration after half revolution is
1 \(\frac{2 V^{2}}{\pi R}\)
2 \(\frac{2 \pi}{R V^{2}}\)
3 \(\frac{2 \mathrm{~V}}{\pi \mathrm{R}^{2}}\)
4 \(\frac{2 \mathrm{R}}{\pi \mathrm{V}}\)
Explanation:
A In a half revolution, change in momentum, \(\Delta \mathrm{P}=\mathrm{mV}-\mathrm{m}(-\mathrm{V})=2 \mathrm{mV}\) The amount of time it takes for a particle to complete a half revolution, \(\mathrm{t}=\frac{\pi \mathrm{R}}{\mathrm{V}}\) Average force \((F)=\) Rate of change in momentum \(=\frac{\Delta \mathrm{P}}{\mathrm{t}}\) \(=\frac{2 \mathrm{mV}}{\frac{\pi \mathrm{R}}{\mathrm{V}}} \quad \text { [From equation (i) and (ii)] }\) \(\mathrm{F}=\frac{2 m \mathrm{~V}^{2}}{\pi \mathrm{R}} \quad\{\because \mathrm{F}=\mathrm{ma}\}\) \(\mathrm{ma}=\frac{2 \mathrm{mV} \mathrm{V}^{2}}{\pi \mathrm{R}} \quad\) \(\mathrm{a}=\frac{2 \mathrm{~V}^{2}}{\pi \mathrm{R}}\)
MHT-CET 2020
Motion in Plane
144035
A particle of mass ' \(m\) ' is performing U.C.M. along a circle of radius ' \(r\) '. The relation between centripetal acceleration ' \(a\) ' and kinetic energy ' \(E\) ' is given by
C We know that \(\mathrm{K} . \mathrm{E}=\frac{1}{2} \mathrm{I} \omega^{2}\) and \(\quad \omega^{2}=\frac{\mathrm{a}}{\mathrm{r}}\) from eqn (i) and (ii) \(\therefore \quad \mathrm{E}=\frac{\mathrm{I}}{2} \times \frac{\mathrm{a}}{\mathrm{r}}\) \(\mathrm{E}=\frac{1}{2} \mathrm{mr}^{2} \times \frac{\mathrm{a}}{\mathrm{r}} \quad\left[\mathrm{I}=\mathrm{mr}^{2}\right]\) \(\mathrm{E}=\frac{1}{2} \mathrm{mr} \cdot \mathrm{a}\) \(\mathrm{a}=\frac{2 \mathrm{E}}{\mathrm{mr}}\)
MHT-CET 2020
Motion in Plane
144036
A train has to negotiate a curve of radius ' \(r\) ' \(m\), the distance between the rails is ' \(\ell\) ' \(\mathrm{m}\) and outer rail is raised above inner rail by distance of ' \(h\) ' \(m\). If the angle of banking is small, the safety speed limit on this banked road is
