149832
A uniform disc of mass \(M\) and radius \(R\) is hinged at its centre \(C\). A force \(F\) is applied on the disc as shown. At this instant, angular acceleration of the disc is
B For Disc, \(M=\) mass of disc \(\mathrm{R}=\) Radius of disc \(\mathrm{F}=\) Applied Force Ans: c : Given that, \(\omega_{0}=0\) \(\mathrm{N}=1800 \mathrm{rpm}\) So, \(\mathrm{t}=2 \mathrm{~min}=120 \mathrm{sec}\) \(\omega=\frac{2 \pi \mathrm{N}}{60}=\frac{2 \pi \times 1800}{60}=60 \pi\) Applying first equation of rotational motion - \(\omega=\omega_{0}+\alpha \times \mathrm{t}\) \(\omega=\alpha \times \mathrm{t} {\left[\because \omega_{0}=0\right]}\) \(\alpha=\frac{\omega}{\mathrm{t}} \) \(\alpha=\frac{60 \pi}{120} {[\because \omega=60 \pi]}\) \(\alpha=\frac{\pi}{2} \mathrm{rad} / \mathrm{sec}^{2}\)
BCECE-2017] We know that
Rotational Motion
149834
A particle describes uniform circular motion in a circle of radius \(2 \mathrm{~m}\), with the angular speed of \(2 \mathrm{rad} \mathrm{s}^{-1}\). The magnitude of the change in its velocity in \(\frac{\pi}{2} \mathrm{~s}\) is
1 \(0 \mathrm{~ms}^{-1}\)
2 \(2 \sqrt{2} \mathrm{~ms}^{-1}\)
3 \(8 \mathrm{~ms}^{-1}\)
4 \(4 \mathrm{~ms}^{-1}\)
5 \(4 \sqrt{2} \mathrm{~ms}^{-1}\)
Explanation:
C Given that, Angular speed \((\omega)=2 \mathrm{rad} / \mathrm{sec}\) Radius \((\mathrm{r})=2 \mathrm{~m}\) Time \((\mathrm{t})=\frac{\pi}{2}\) sec Angle moved in time \(\mathrm{t}\), Now, \(\begin{aligned} \text { Angular velocity }(\mathrm{v}) =\omega \mathrm{r} \\ =2 \times 2\end{aligned}\) \(=2 \times 2\) \(\mathrm{v} =4 \mathrm{~m} / \mathrm{sec}\) The magnitude of the change in velocity, \(\Delta \mathrm{v}=2 \mathrm{v} \sin \frac{\theta}{2}\) \(\Delta \mathrm{v}=2 \times 4 \times \sin \frac{\pi}{2}\) \(\Delta \mathrm{v}=8 \times 1\) \(\Delta \mathrm{v}=8 \mathrm{~m} / \mathrm{sec}\)
Kerala CEE- 2013
Rotational Motion
149835
The angular velocity of a wheel increases from \(100 \mathrm{rps}\) to \(300 \mathrm{rps}\) in \(10 \mathrm{~s}\). The number of revolutions made during that time is
1 600
2 1500
3 1000
4 3000
5 2000,
Explanation:
E Given that, \(\mathrm{N}_{1}=100 \mathrm{rps}, \quad \mathrm{N}_{2}=300 \mathrm{rps}\) \(\mathrm{t}=10 \mathrm{sec}\) We know, \(\omega=2 \pi \mathrm{N}\) \(\omega_{\mathrm{i}}=2 \pi \times 100=200 \pi\) \(\omega_{\mathrm{f}}=2 \pi \times 300=600 \pi\) Now, \(\quad \omega_{\mathrm{f}}=\omega_{\mathrm{i}}+\alpha \mathrm{t}\) \(600 \pi=200 \pi+10 \alpha\) \(400 \pi=10 \alpha\) \(\alpha=40 \pi\) From equation of motion- \(\left(\omega_{\mathrm{f}}\right)^{2}=\left(\omega_{\mathrm{i}}\right)^{2}+2 \alpha \theta\) So, \((600 \pi)^{2}=(200 \pi)^{2}+2 \times 40 \pi \times \theta\) \(\pi^{2}(360000-40000)=80 \pi \theta\) \(320000 \pi=80 \theta\) \(4000 \pi=\theta\) So, Number of revolutions \(=\frac{\theta}{1 \text { revolution angle }(2 \pi)}\) \(=\frac{4000 \pi}{2 \pi}=2000\)
Kerala CEE - 2008
Rotational Motion
149836
A particle of mass \(m=5\) units is moving with a uniform speed \(v=3 \sqrt{2} \mathrm{~m}\) in the XOY plane along the line \(Y=X+4\). The magnitude of the angular momentum about origin is :
1 zero
2 60 unit
3 7.5 unit
4 \(40 \sqrt{2}\) unit
5 3.0 unit
Explanation:
B Given that, \(\mathrm{v}=3 \sqrt{2} \mathrm{~m}\) \(Y=X+4\) Comparing above equation with \(\mathrm{Y}=\tan \theta \mathrm{X}+\mathrm{c}\), \(\tan \theta=1\) \(\theta=45^{\circ}\) We know that, \(\mathrm{L}=\mathrm{mvr}_{\perp}\) Now, from diagram, \(\sin 45^{\circ}=\frac{x}{4}\) \(\mathrm{x}=4 \sin 45^{\circ}\) \(x=4 \times \frac{1}{\sqrt{2}}\) So, \(x=2 \sqrt{2}\) \(\mathrm{L} =\mathrm{mv} \times \mathrm{x}\) \(\mathrm{L} =5 \times 3 \sqrt{2} \times 2 \sqrt{2} \quad\left\{\mathrm{r}_{\perp}=\mathrm{x}\right\}\) \(\mathrm{L} =60 \text { unit }\)
149832
A uniform disc of mass \(M\) and radius \(R\) is hinged at its centre \(C\). A force \(F\) is applied on the disc as shown. At this instant, angular acceleration of the disc is
B For Disc, \(M=\) mass of disc \(\mathrm{R}=\) Radius of disc \(\mathrm{F}=\) Applied Force Ans: c : Given that, \(\omega_{0}=0\) \(\mathrm{N}=1800 \mathrm{rpm}\) So, \(\mathrm{t}=2 \mathrm{~min}=120 \mathrm{sec}\) \(\omega=\frac{2 \pi \mathrm{N}}{60}=\frac{2 \pi \times 1800}{60}=60 \pi\) Applying first equation of rotational motion - \(\omega=\omega_{0}+\alpha \times \mathrm{t}\) \(\omega=\alpha \times \mathrm{t} {\left[\because \omega_{0}=0\right]}\) \(\alpha=\frac{\omega}{\mathrm{t}} \) \(\alpha=\frac{60 \pi}{120} {[\because \omega=60 \pi]}\) \(\alpha=\frac{\pi}{2} \mathrm{rad} / \mathrm{sec}^{2}\)
BCECE-2017] We know that
Rotational Motion
149834
A particle describes uniform circular motion in a circle of radius \(2 \mathrm{~m}\), with the angular speed of \(2 \mathrm{rad} \mathrm{s}^{-1}\). The magnitude of the change in its velocity in \(\frac{\pi}{2} \mathrm{~s}\) is
1 \(0 \mathrm{~ms}^{-1}\)
2 \(2 \sqrt{2} \mathrm{~ms}^{-1}\)
3 \(8 \mathrm{~ms}^{-1}\)
4 \(4 \mathrm{~ms}^{-1}\)
5 \(4 \sqrt{2} \mathrm{~ms}^{-1}\)
Explanation:
C Given that, Angular speed \((\omega)=2 \mathrm{rad} / \mathrm{sec}\) Radius \((\mathrm{r})=2 \mathrm{~m}\) Time \((\mathrm{t})=\frac{\pi}{2}\) sec Angle moved in time \(\mathrm{t}\), Now, \(\begin{aligned} \text { Angular velocity }(\mathrm{v}) =\omega \mathrm{r} \\ =2 \times 2\end{aligned}\) \(=2 \times 2\) \(\mathrm{v} =4 \mathrm{~m} / \mathrm{sec}\) The magnitude of the change in velocity, \(\Delta \mathrm{v}=2 \mathrm{v} \sin \frac{\theta}{2}\) \(\Delta \mathrm{v}=2 \times 4 \times \sin \frac{\pi}{2}\) \(\Delta \mathrm{v}=8 \times 1\) \(\Delta \mathrm{v}=8 \mathrm{~m} / \mathrm{sec}\)
Kerala CEE- 2013
Rotational Motion
149835
The angular velocity of a wheel increases from \(100 \mathrm{rps}\) to \(300 \mathrm{rps}\) in \(10 \mathrm{~s}\). The number of revolutions made during that time is
1 600
2 1500
3 1000
4 3000
5 2000,
Explanation:
E Given that, \(\mathrm{N}_{1}=100 \mathrm{rps}, \quad \mathrm{N}_{2}=300 \mathrm{rps}\) \(\mathrm{t}=10 \mathrm{sec}\) We know, \(\omega=2 \pi \mathrm{N}\) \(\omega_{\mathrm{i}}=2 \pi \times 100=200 \pi\) \(\omega_{\mathrm{f}}=2 \pi \times 300=600 \pi\) Now, \(\quad \omega_{\mathrm{f}}=\omega_{\mathrm{i}}+\alpha \mathrm{t}\) \(600 \pi=200 \pi+10 \alpha\) \(400 \pi=10 \alpha\) \(\alpha=40 \pi\) From equation of motion- \(\left(\omega_{\mathrm{f}}\right)^{2}=\left(\omega_{\mathrm{i}}\right)^{2}+2 \alpha \theta\) So, \((600 \pi)^{2}=(200 \pi)^{2}+2 \times 40 \pi \times \theta\) \(\pi^{2}(360000-40000)=80 \pi \theta\) \(320000 \pi=80 \theta\) \(4000 \pi=\theta\) So, Number of revolutions \(=\frac{\theta}{1 \text { revolution angle }(2 \pi)}\) \(=\frac{4000 \pi}{2 \pi}=2000\)
Kerala CEE - 2008
Rotational Motion
149836
A particle of mass \(m=5\) units is moving with a uniform speed \(v=3 \sqrt{2} \mathrm{~m}\) in the XOY plane along the line \(Y=X+4\). The magnitude of the angular momentum about origin is :
1 zero
2 60 unit
3 7.5 unit
4 \(40 \sqrt{2}\) unit
5 3.0 unit
Explanation:
B Given that, \(\mathrm{v}=3 \sqrt{2} \mathrm{~m}\) \(Y=X+4\) Comparing above equation with \(\mathrm{Y}=\tan \theta \mathrm{X}+\mathrm{c}\), \(\tan \theta=1\) \(\theta=45^{\circ}\) We know that, \(\mathrm{L}=\mathrm{mvr}_{\perp}\) Now, from diagram, \(\sin 45^{\circ}=\frac{x}{4}\) \(\mathrm{x}=4 \sin 45^{\circ}\) \(x=4 \times \frac{1}{\sqrt{2}}\) So, \(x=2 \sqrt{2}\) \(\mathrm{L} =\mathrm{mv} \times \mathrm{x}\) \(\mathrm{L} =5 \times 3 \sqrt{2} \times 2 \sqrt{2} \quad\left\{\mathrm{r}_{\perp}=\mathrm{x}\right\}\) \(\mathrm{L} =60 \text { unit }\)
149832
A uniform disc of mass \(M\) and radius \(R\) is hinged at its centre \(C\). A force \(F\) is applied on the disc as shown. At this instant, angular acceleration of the disc is
B For Disc, \(M=\) mass of disc \(\mathrm{R}=\) Radius of disc \(\mathrm{F}=\) Applied Force Ans: c : Given that, \(\omega_{0}=0\) \(\mathrm{N}=1800 \mathrm{rpm}\) So, \(\mathrm{t}=2 \mathrm{~min}=120 \mathrm{sec}\) \(\omega=\frac{2 \pi \mathrm{N}}{60}=\frac{2 \pi \times 1800}{60}=60 \pi\) Applying first equation of rotational motion - \(\omega=\omega_{0}+\alpha \times \mathrm{t}\) \(\omega=\alpha \times \mathrm{t} {\left[\because \omega_{0}=0\right]}\) \(\alpha=\frac{\omega}{\mathrm{t}} \) \(\alpha=\frac{60 \pi}{120} {[\because \omega=60 \pi]}\) \(\alpha=\frac{\pi}{2} \mathrm{rad} / \mathrm{sec}^{2}\)
BCECE-2017] We know that
Rotational Motion
149834
A particle describes uniform circular motion in a circle of radius \(2 \mathrm{~m}\), with the angular speed of \(2 \mathrm{rad} \mathrm{s}^{-1}\). The magnitude of the change in its velocity in \(\frac{\pi}{2} \mathrm{~s}\) is
1 \(0 \mathrm{~ms}^{-1}\)
2 \(2 \sqrt{2} \mathrm{~ms}^{-1}\)
3 \(8 \mathrm{~ms}^{-1}\)
4 \(4 \mathrm{~ms}^{-1}\)
5 \(4 \sqrt{2} \mathrm{~ms}^{-1}\)
Explanation:
C Given that, Angular speed \((\omega)=2 \mathrm{rad} / \mathrm{sec}\) Radius \((\mathrm{r})=2 \mathrm{~m}\) Time \((\mathrm{t})=\frac{\pi}{2}\) sec Angle moved in time \(\mathrm{t}\), Now, \(\begin{aligned} \text { Angular velocity }(\mathrm{v}) =\omega \mathrm{r} \\ =2 \times 2\end{aligned}\) \(=2 \times 2\) \(\mathrm{v} =4 \mathrm{~m} / \mathrm{sec}\) The magnitude of the change in velocity, \(\Delta \mathrm{v}=2 \mathrm{v} \sin \frac{\theta}{2}\) \(\Delta \mathrm{v}=2 \times 4 \times \sin \frac{\pi}{2}\) \(\Delta \mathrm{v}=8 \times 1\) \(\Delta \mathrm{v}=8 \mathrm{~m} / \mathrm{sec}\)
Kerala CEE- 2013
Rotational Motion
149835
The angular velocity of a wheel increases from \(100 \mathrm{rps}\) to \(300 \mathrm{rps}\) in \(10 \mathrm{~s}\). The number of revolutions made during that time is
1 600
2 1500
3 1000
4 3000
5 2000,
Explanation:
E Given that, \(\mathrm{N}_{1}=100 \mathrm{rps}, \quad \mathrm{N}_{2}=300 \mathrm{rps}\) \(\mathrm{t}=10 \mathrm{sec}\) We know, \(\omega=2 \pi \mathrm{N}\) \(\omega_{\mathrm{i}}=2 \pi \times 100=200 \pi\) \(\omega_{\mathrm{f}}=2 \pi \times 300=600 \pi\) Now, \(\quad \omega_{\mathrm{f}}=\omega_{\mathrm{i}}+\alpha \mathrm{t}\) \(600 \pi=200 \pi+10 \alpha\) \(400 \pi=10 \alpha\) \(\alpha=40 \pi\) From equation of motion- \(\left(\omega_{\mathrm{f}}\right)^{2}=\left(\omega_{\mathrm{i}}\right)^{2}+2 \alpha \theta\) So, \((600 \pi)^{2}=(200 \pi)^{2}+2 \times 40 \pi \times \theta\) \(\pi^{2}(360000-40000)=80 \pi \theta\) \(320000 \pi=80 \theta\) \(4000 \pi=\theta\) So, Number of revolutions \(=\frac{\theta}{1 \text { revolution angle }(2 \pi)}\) \(=\frac{4000 \pi}{2 \pi}=2000\)
Kerala CEE - 2008
Rotational Motion
149836
A particle of mass \(m=5\) units is moving with a uniform speed \(v=3 \sqrt{2} \mathrm{~m}\) in the XOY plane along the line \(Y=X+4\). The magnitude of the angular momentum about origin is :
1 zero
2 60 unit
3 7.5 unit
4 \(40 \sqrt{2}\) unit
5 3.0 unit
Explanation:
B Given that, \(\mathrm{v}=3 \sqrt{2} \mathrm{~m}\) \(Y=X+4\) Comparing above equation with \(\mathrm{Y}=\tan \theta \mathrm{X}+\mathrm{c}\), \(\tan \theta=1\) \(\theta=45^{\circ}\) We know that, \(\mathrm{L}=\mathrm{mvr}_{\perp}\) Now, from diagram, \(\sin 45^{\circ}=\frac{x}{4}\) \(\mathrm{x}=4 \sin 45^{\circ}\) \(x=4 \times \frac{1}{\sqrt{2}}\) So, \(x=2 \sqrt{2}\) \(\mathrm{L} =\mathrm{mv} \times \mathrm{x}\) \(\mathrm{L} =5 \times 3 \sqrt{2} \times 2 \sqrt{2} \quad\left\{\mathrm{r}_{\perp}=\mathrm{x}\right\}\) \(\mathrm{L} =60 \text { unit }\)
149832
A uniform disc of mass \(M\) and radius \(R\) is hinged at its centre \(C\). A force \(F\) is applied on the disc as shown. At this instant, angular acceleration of the disc is
B For Disc, \(M=\) mass of disc \(\mathrm{R}=\) Radius of disc \(\mathrm{F}=\) Applied Force Ans: c : Given that, \(\omega_{0}=0\) \(\mathrm{N}=1800 \mathrm{rpm}\) So, \(\mathrm{t}=2 \mathrm{~min}=120 \mathrm{sec}\) \(\omega=\frac{2 \pi \mathrm{N}}{60}=\frac{2 \pi \times 1800}{60}=60 \pi\) Applying first equation of rotational motion - \(\omega=\omega_{0}+\alpha \times \mathrm{t}\) \(\omega=\alpha \times \mathrm{t} {\left[\because \omega_{0}=0\right]}\) \(\alpha=\frac{\omega}{\mathrm{t}} \) \(\alpha=\frac{60 \pi}{120} {[\because \omega=60 \pi]}\) \(\alpha=\frac{\pi}{2} \mathrm{rad} / \mathrm{sec}^{2}\)
BCECE-2017] We know that
Rotational Motion
149834
A particle describes uniform circular motion in a circle of radius \(2 \mathrm{~m}\), with the angular speed of \(2 \mathrm{rad} \mathrm{s}^{-1}\). The magnitude of the change in its velocity in \(\frac{\pi}{2} \mathrm{~s}\) is
1 \(0 \mathrm{~ms}^{-1}\)
2 \(2 \sqrt{2} \mathrm{~ms}^{-1}\)
3 \(8 \mathrm{~ms}^{-1}\)
4 \(4 \mathrm{~ms}^{-1}\)
5 \(4 \sqrt{2} \mathrm{~ms}^{-1}\)
Explanation:
C Given that, Angular speed \((\omega)=2 \mathrm{rad} / \mathrm{sec}\) Radius \((\mathrm{r})=2 \mathrm{~m}\) Time \((\mathrm{t})=\frac{\pi}{2}\) sec Angle moved in time \(\mathrm{t}\), Now, \(\begin{aligned} \text { Angular velocity }(\mathrm{v}) =\omega \mathrm{r} \\ =2 \times 2\end{aligned}\) \(=2 \times 2\) \(\mathrm{v} =4 \mathrm{~m} / \mathrm{sec}\) The magnitude of the change in velocity, \(\Delta \mathrm{v}=2 \mathrm{v} \sin \frac{\theta}{2}\) \(\Delta \mathrm{v}=2 \times 4 \times \sin \frac{\pi}{2}\) \(\Delta \mathrm{v}=8 \times 1\) \(\Delta \mathrm{v}=8 \mathrm{~m} / \mathrm{sec}\)
Kerala CEE- 2013
Rotational Motion
149835
The angular velocity of a wheel increases from \(100 \mathrm{rps}\) to \(300 \mathrm{rps}\) in \(10 \mathrm{~s}\). The number of revolutions made during that time is
1 600
2 1500
3 1000
4 3000
5 2000,
Explanation:
E Given that, \(\mathrm{N}_{1}=100 \mathrm{rps}, \quad \mathrm{N}_{2}=300 \mathrm{rps}\) \(\mathrm{t}=10 \mathrm{sec}\) We know, \(\omega=2 \pi \mathrm{N}\) \(\omega_{\mathrm{i}}=2 \pi \times 100=200 \pi\) \(\omega_{\mathrm{f}}=2 \pi \times 300=600 \pi\) Now, \(\quad \omega_{\mathrm{f}}=\omega_{\mathrm{i}}+\alpha \mathrm{t}\) \(600 \pi=200 \pi+10 \alpha\) \(400 \pi=10 \alpha\) \(\alpha=40 \pi\) From equation of motion- \(\left(\omega_{\mathrm{f}}\right)^{2}=\left(\omega_{\mathrm{i}}\right)^{2}+2 \alpha \theta\) So, \((600 \pi)^{2}=(200 \pi)^{2}+2 \times 40 \pi \times \theta\) \(\pi^{2}(360000-40000)=80 \pi \theta\) \(320000 \pi=80 \theta\) \(4000 \pi=\theta\) So, Number of revolutions \(=\frac{\theta}{1 \text { revolution angle }(2 \pi)}\) \(=\frac{4000 \pi}{2 \pi}=2000\)
Kerala CEE - 2008
Rotational Motion
149836
A particle of mass \(m=5\) units is moving with a uniform speed \(v=3 \sqrt{2} \mathrm{~m}\) in the XOY plane along the line \(Y=X+4\). The magnitude of the angular momentum about origin is :
1 zero
2 60 unit
3 7.5 unit
4 \(40 \sqrt{2}\) unit
5 3.0 unit
Explanation:
B Given that, \(\mathrm{v}=3 \sqrt{2} \mathrm{~m}\) \(Y=X+4\) Comparing above equation with \(\mathrm{Y}=\tan \theta \mathrm{X}+\mathrm{c}\), \(\tan \theta=1\) \(\theta=45^{\circ}\) We know that, \(\mathrm{L}=\mathrm{mvr}_{\perp}\) Now, from diagram, \(\sin 45^{\circ}=\frac{x}{4}\) \(\mathrm{x}=4 \sin 45^{\circ}\) \(x=4 \times \frac{1}{\sqrt{2}}\) So, \(x=2 \sqrt{2}\) \(\mathrm{L} =\mathrm{mv} \times \mathrm{x}\) \(\mathrm{L} =5 \times 3 \sqrt{2} \times 2 \sqrt{2} \quad\left\{\mathrm{r}_{\perp}=\mathrm{x}\right\}\) \(\mathrm{L} =60 \text { unit }\)
