150442
A body is rolling down an inclined plane. Its translational and rotational kinetic energies are equal. The body is a :
1 Solid sphere
2 Hollow sphere
3 Solid cylinder
4 Hollow cylinder
Explanation:
D The body is rolling down an inclined plane, then its translational and rotational kinetic energies both are equal. Rotational kinetic energy \(\left(K_{R}\right)=\frac{1}{2} I \omega^{2}\) Where, moment of inertia \(=\mathrm{I}\) Angular velocity \(=\omega\) Translational kinetic energy \(\left(\mathrm{K}_{\mathrm{T}}\right)=\frac{1}{2} \mathrm{mv}^{2}\) \(\mathrm{K}_{\mathrm{T}}=\frac{1}{2} \mathrm{~m}(\mathrm{r} \omega)^{2} \quad \ldots .(\mathrm{i}) \quad(\because \mathrm{v}=\mathrm{r} \omega\) Solid sphere - moment of inertia of solid sphere \(\mathrm{I}=\frac{2}{5} \mathrm{mr}^{2}\) \(\therefore \quad \mathrm{K}_{\mathrm{R}}=\frac{1}{2} \mathrm{I} \omega^{2}\) \(\mathrm{K}_{\mathrm{R}}=\frac{1}{2} \cdot \frac{2}{5} \mathrm{mr}^{2} \cdot \omega^{2}=\frac{1}{5} \mathrm{mr}^{2} \omega^{2}\) So, \(\mathrm{K}_{\mathrm{R}} \neq \mathrm{K}_{\mathrm{T}}\) From eqn. (i) Solid cylinder - moment of inertia of solid cylinder \(\mathrm{I}=\frac{1}{2} \mathrm{mr}^{2}\) \(\mathrm{K}_{\mathrm{R}}=\frac{1}{2}\left(\frac{1}{2} \mathrm{mr}^{2}\right) \omega^{2}\) \(\mathrm{K}_{\mathrm{R}} =\frac{1}{4} \mathrm{mr}^{2} \omega^{2}\) \(\mathrm{~K}_{\mathrm{R}} \neq \mathrm{K}_{\mathrm{T}}\) Hollow cylinder- moment of inertia of hollow cylinder is \(\therefore \quad \mathrm{I} =\mathrm{mr}^{2}\) \(\quad \mathrm{~K}_{\mathrm{R}} =\frac{1}{2} \mathrm{I} \omega^{2}\) \(\mathrm{~K}_{\mathrm{R}} =\frac{1}{2} \mathrm{mr}^{2} \omega^{2}\) So, \(\mathrm{K}_{\mathrm{R}}=\mathrm{K}_{\mathrm{T}}\) So, option (d) is correct answer.
UPSEE - 2004
Rotational Motion
150444
One solid sphere and disc of same radius are falling along an inclined plane without slipping. One reaches earlier than the other due to
1 different radius of gyration
2 different sizes
3 different friction
4 different moment of inertia
Explanation:
A (i) For solid sphere, the moment of inertia about the diameter is \(\mathrm{I}_{\mathrm{s}}=\frac{2}{5} \mathrm{MR}^{2}\) Now \(\mathrm{I}=\mathrm{MK}^{2}\) for anybody, where \(\mathrm{K}\) is radius of gyration of that body. So \(\mathrm{MK}^{2}=\frac{2}{5} \mathrm{MR}^{2} \Rightarrow \mathrm{K}=\mathrm{R} \sqrt{2 / 5}\) (ii) The moment of inertia of disc about an axis passing through its centre \ perpendicular to plane is \(I_{d} =\frac{M R^{2}}{2}=M K^{2}\) \(K =R \sqrt{\frac{1}{2}}\) Now acceleration of anybody which is rolling on an inclined plane is \(a=\left(\frac{g \sin \theta}{1+K^{2} / R^{2}}\right)\) For same R, \(\mathrm{a} \propto \frac{1}{\mathrm{~K}^{2}}\) \(\Rightarrow \quad \mathrm{a}_{\mathrm{s}}>\mathrm{a}_{\mathrm{d}}\) So, solid sphere will reach earlier to bottom of an inclined plane than disc.
BITSAT-2008
Rotational Motion
150445
The ratio of the accelerations for a solid sphere (mass ' \(m\) ' and radius ' \(R\) ') rolling down an incline of angle ' \(\theta\) ' without slipping and slipping down the incline without rolling is :
1 \(5: 7\)
2 \(2: 3\)
3 \(2: 5\)
4 \(7: 5\)
Explanation:
A For solid sphere rolling without slipping on inclined plane, acceleration \(a_{1}=\frac{g \sin \theta}{\left(1+\frac{K^{2}}{R^{2}}\right)}\) For solid sphere slipping on inclined plane without rolling, acceleration \(a_{2}=g \sin \theta\) For solid sphere, \(\mathrm{I}=\frac{2}{5} \mathrm{MR}^{2}\) \(\mathrm{MK}^{2} =\frac{2}{5} \mathrm{MR}^{2}\) \(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}} =\frac{2}{5}\) Therefore, required ratio, \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{1}{\left(1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)}=\frac{1}{\left(1+\frac{2}{5}\right)}=\frac{5}{7}\) \(\text { i.e. } \quad a_{1}: a_{2}=5: 7\)
AIPMT-2014
Rotational Motion
150446
A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in figure. At any instant, for the lower most point of the disc-
1 velocity is \(\mathrm{v}\), acceleration is zero
2 velocity is zero, acceleration is zero
3 velocity is \(v\), acceleration is \(v^{2} / R\)
4 velocity is zero, acceleration is \(v^{2} / R\)
Explanation:
D Velocity at point A is 0 (zero) because disc performing pure rolling. Acceleration at point \(\mathrm{A}\) is, We know that \(\begin{aligned} \mathrm{v} =\mathrm{R} \omega \\ \text { Or } \quad \omega =\frac{\mathrm{v}}{\mathrm{R}}\end{aligned}\) Angular acceleration is \(\alpha)=R \omega^{2}\) \(\alpha=\mathrm{R} \cdot \frac{\mathrm{v}^{2}}{\mathrm{R}^{2}}=\frac{\mathrm{v}^{2}}{\mathrm{R}}\) So, Acceleration at lower most point of disc is \(\frac{\mathrm{v}^{2}}{\mathrm{R}}\).
150442
A body is rolling down an inclined plane. Its translational and rotational kinetic energies are equal. The body is a :
1 Solid sphere
2 Hollow sphere
3 Solid cylinder
4 Hollow cylinder
Explanation:
D The body is rolling down an inclined plane, then its translational and rotational kinetic energies both are equal. Rotational kinetic energy \(\left(K_{R}\right)=\frac{1}{2} I \omega^{2}\) Where, moment of inertia \(=\mathrm{I}\) Angular velocity \(=\omega\) Translational kinetic energy \(\left(\mathrm{K}_{\mathrm{T}}\right)=\frac{1}{2} \mathrm{mv}^{2}\) \(\mathrm{K}_{\mathrm{T}}=\frac{1}{2} \mathrm{~m}(\mathrm{r} \omega)^{2} \quad \ldots .(\mathrm{i}) \quad(\because \mathrm{v}=\mathrm{r} \omega\) Solid sphere - moment of inertia of solid sphere \(\mathrm{I}=\frac{2}{5} \mathrm{mr}^{2}\) \(\therefore \quad \mathrm{K}_{\mathrm{R}}=\frac{1}{2} \mathrm{I} \omega^{2}\) \(\mathrm{K}_{\mathrm{R}}=\frac{1}{2} \cdot \frac{2}{5} \mathrm{mr}^{2} \cdot \omega^{2}=\frac{1}{5} \mathrm{mr}^{2} \omega^{2}\) So, \(\mathrm{K}_{\mathrm{R}} \neq \mathrm{K}_{\mathrm{T}}\) From eqn. (i) Solid cylinder - moment of inertia of solid cylinder \(\mathrm{I}=\frac{1}{2} \mathrm{mr}^{2}\) \(\mathrm{K}_{\mathrm{R}}=\frac{1}{2}\left(\frac{1}{2} \mathrm{mr}^{2}\right) \omega^{2}\) \(\mathrm{K}_{\mathrm{R}} =\frac{1}{4} \mathrm{mr}^{2} \omega^{2}\) \(\mathrm{~K}_{\mathrm{R}} \neq \mathrm{K}_{\mathrm{T}}\) Hollow cylinder- moment of inertia of hollow cylinder is \(\therefore \quad \mathrm{I} =\mathrm{mr}^{2}\) \(\quad \mathrm{~K}_{\mathrm{R}} =\frac{1}{2} \mathrm{I} \omega^{2}\) \(\mathrm{~K}_{\mathrm{R}} =\frac{1}{2} \mathrm{mr}^{2} \omega^{2}\) So, \(\mathrm{K}_{\mathrm{R}}=\mathrm{K}_{\mathrm{T}}\) So, option (d) is correct answer.
UPSEE - 2004
Rotational Motion
150444
One solid sphere and disc of same radius are falling along an inclined plane without slipping. One reaches earlier than the other due to
1 different radius of gyration
2 different sizes
3 different friction
4 different moment of inertia
Explanation:
A (i) For solid sphere, the moment of inertia about the diameter is \(\mathrm{I}_{\mathrm{s}}=\frac{2}{5} \mathrm{MR}^{2}\) Now \(\mathrm{I}=\mathrm{MK}^{2}\) for anybody, where \(\mathrm{K}\) is radius of gyration of that body. So \(\mathrm{MK}^{2}=\frac{2}{5} \mathrm{MR}^{2} \Rightarrow \mathrm{K}=\mathrm{R} \sqrt{2 / 5}\) (ii) The moment of inertia of disc about an axis passing through its centre \ perpendicular to plane is \(I_{d} =\frac{M R^{2}}{2}=M K^{2}\) \(K =R \sqrt{\frac{1}{2}}\) Now acceleration of anybody which is rolling on an inclined plane is \(a=\left(\frac{g \sin \theta}{1+K^{2} / R^{2}}\right)\) For same R, \(\mathrm{a} \propto \frac{1}{\mathrm{~K}^{2}}\) \(\Rightarrow \quad \mathrm{a}_{\mathrm{s}}>\mathrm{a}_{\mathrm{d}}\) So, solid sphere will reach earlier to bottom of an inclined plane than disc.
BITSAT-2008
Rotational Motion
150445
The ratio of the accelerations for a solid sphere (mass ' \(m\) ' and radius ' \(R\) ') rolling down an incline of angle ' \(\theta\) ' without slipping and slipping down the incline without rolling is :
1 \(5: 7\)
2 \(2: 3\)
3 \(2: 5\)
4 \(7: 5\)
Explanation:
A For solid sphere rolling without slipping on inclined plane, acceleration \(a_{1}=\frac{g \sin \theta}{\left(1+\frac{K^{2}}{R^{2}}\right)}\) For solid sphere slipping on inclined plane without rolling, acceleration \(a_{2}=g \sin \theta\) For solid sphere, \(\mathrm{I}=\frac{2}{5} \mathrm{MR}^{2}\) \(\mathrm{MK}^{2} =\frac{2}{5} \mathrm{MR}^{2}\) \(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}} =\frac{2}{5}\) Therefore, required ratio, \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{1}{\left(1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)}=\frac{1}{\left(1+\frac{2}{5}\right)}=\frac{5}{7}\) \(\text { i.e. } \quad a_{1}: a_{2}=5: 7\)
AIPMT-2014
Rotational Motion
150446
A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in figure. At any instant, for the lower most point of the disc-
1 velocity is \(\mathrm{v}\), acceleration is zero
2 velocity is zero, acceleration is zero
3 velocity is \(v\), acceleration is \(v^{2} / R\)
4 velocity is zero, acceleration is \(v^{2} / R\)
Explanation:
D Velocity at point A is 0 (zero) because disc performing pure rolling. Acceleration at point \(\mathrm{A}\) is, We know that \(\begin{aligned} \mathrm{v} =\mathrm{R} \omega \\ \text { Or } \quad \omega =\frac{\mathrm{v}}{\mathrm{R}}\end{aligned}\) Angular acceleration is \(\alpha)=R \omega^{2}\) \(\alpha=\mathrm{R} \cdot \frac{\mathrm{v}^{2}}{\mathrm{R}^{2}}=\frac{\mathrm{v}^{2}}{\mathrm{R}}\) So, Acceleration at lower most point of disc is \(\frac{\mathrm{v}^{2}}{\mathrm{R}}\).
150442
A body is rolling down an inclined plane. Its translational and rotational kinetic energies are equal. The body is a :
1 Solid sphere
2 Hollow sphere
3 Solid cylinder
4 Hollow cylinder
Explanation:
D The body is rolling down an inclined plane, then its translational and rotational kinetic energies both are equal. Rotational kinetic energy \(\left(K_{R}\right)=\frac{1}{2} I \omega^{2}\) Where, moment of inertia \(=\mathrm{I}\) Angular velocity \(=\omega\) Translational kinetic energy \(\left(\mathrm{K}_{\mathrm{T}}\right)=\frac{1}{2} \mathrm{mv}^{2}\) \(\mathrm{K}_{\mathrm{T}}=\frac{1}{2} \mathrm{~m}(\mathrm{r} \omega)^{2} \quad \ldots .(\mathrm{i}) \quad(\because \mathrm{v}=\mathrm{r} \omega\) Solid sphere - moment of inertia of solid sphere \(\mathrm{I}=\frac{2}{5} \mathrm{mr}^{2}\) \(\therefore \quad \mathrm{K}_{\mathrm{R}}=\frac{1}{2} \mathrm{I} \omega^{2}\) \(\mathrm{K}_{\mathrm{R}}=\frac{1}{2} \cdot \frac{2}{5} \mathrm{mr}^{2} \cdot \omega^{2}=\frac{1}{5} \mathrm{mr}^{2} \omega^{2}\) So, \(\mathrm{K}_{\mathrm{R}} \neq \mathrm{K}_{\mathrm{T}}\) From eqn. (i) Solid cylinder - moment of inertia of solid cylinder \(\mathrm{I}=\frac{1}{2} \mathrm{mr}^{2}\) \(\mathrm{K}_{\mathrm{R}}=\frac{1}{2}\left(\frac{1}{2} \mathrm{mr}^{2}\right) \omega^{2}\) \(\mathrm{K}_{\mathrm{R}} =\frac{1}{4} \mathrm{mr}^{2} \omega^{2}\) \(\mathrm{~K}_{\mathrm{R}} \neq \mathrm{K}_{\mathrm{T}}\) Hollow cylinder- moment of inertia of hollow cylinder is \(\therefore \quad \mathrm{I} =\mathrm{mr}^{2}\) \(\quad \mathrm{~K}_{\mathrm{R}} =\frac{1}{2} \mathrm{I} \omega^{2}\) \(\mathrm{~K}_{\mathrm{R}} =\frac{1}{2} \mathrm{mr}^{2} \omega^{2}\) So, \(\mathrm{K}_{\mathrm{R}}=\mathrm{K}_{\mathrm{T}}\) So, option (d) is correct answer.
UPSEE - 2004
Rotational Motion
150444
One solid sphere and disc of same radius are falling along an inclined plane without slipping. One reaches earlier than the other due to
1 different radius of gyration
2 different sizes
3 different friction
4 different moment of inertia
Explanation:
A (i) For solid sphere, the moment of inertia about the diameter is \(\mathrm{I}_{\mathrm{s}}=\frac{2}{5} \mathrm{MR}^{2}\) Now \(\mathrm{I}=\mathrm{MK}^{2}\) for anybody, where \(\mathrm{K}\) is radius of gyration of that body. So \(\mathrm{MK}^{2}=\frac{2}{5} \mathrm{MR}^{2} \Rightarrow \mathrm{K}=\mathrm{R} \sqrt{2 / 5}\) (ii) The moment of inertia of disc about an axis passing through its centre \ perpendicular to plane is \(I_{d} =\frac{M R^{2}}{2}=M K^{2}\) \(K =R \sqrt{\frac{1}{2}}\) Now acceleration of anybody which is rolling on an inclined plane is \(a=\left(\frac{g \sin \theta}{1+K^{2} / R^{2}}\right)\) For same R, \(\mathrm{a} \propto \frac{1}{\mathrm{~K}^{2}}\) \(\Rightarrow \quad \mathrm{a}_{\mathrm{s}}>\mathrm{a}_{\mathrm{d}}\) So, solid sphere will reach earlier to bottom of an inclined plane than disc.
BITSAT-2008
Rotational Motion
150445
The ratio of the accelerations for a solid sphere (mass ' \(m\) ' and radius ' \(R\) ') rolling down an incline of angle ' \(\theta\) ' without slipping and slipping down the incline without rolling is :
1 \(5: 7\)
2 \(2: 3\)
3 \(2: 5\)
4 \(7: 5\)
Explanation:
A For solid sphere rolling without slipping on inclined plane, acceleration \(a_{1}=\frac{g \sin \theta}{\left(1+\frac{K^{2}}{R^{2}}\right)}\) For solid sphere slipping on inclined plane without rolling, acceleration \(a_{2}=g \sin \theta\) For solid sphere, \(\mathrm{I}=\frac{2}{5} \mathrm{MR}^{2}\) \(\mathrm{MK}^{2} =\frac{2}{5} \mathrm{MR}^{2}\) \(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}} =\frac{2}{5}\) Therefore, required ratio, \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{1}{\left(1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)}=\frac{1}{\left(1+\frac{2}{5}\right)}=\frac{5}{7}\) \(\text { i.e. } \quad a_{1}: a_{2}=5: 7\)
AIPMT-2014
Rotational Motion
150446
A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in figure. At any instant, for the lower most point of the disc-
1 velocity is \(\mathrm{v}\), acceleration is zero
2 velocity is zero, acceleration is zero
3 velocity is \(v\), acceleration is \(v^{2} / R\)
4 velocity is zero, acceleration is \(v^{2} / R\)
Explanation:
D Velocity at point A is 0 (zero) because disc performing pure rolling. Acceleration at point \(\mathrm{A}\) is, We know that \(\begin{aligned} \mathrm{v} =\mathrm{R} \omega \\ \text { Or } \quad \omega =\frac{\mathrm{v}}{\mathrm{R}}\end{aligned}\) Angular acceleration is \(\alpha)=R \omega^{2}\) \(\alpha=\mathrm{R} \cdot \frac{\mathrm{v}^{2}}{\mathrm{R}^{2}}=\frac{\mathrm{v}^{2}}{\mathrm{R}}\) So, Acceleration at lower most point of disc is \(\frac{\mathrm{v}^{2}}{\mathrm{R}}\).
NEET Test Series from KOTA - 10 Papers In MS WORD
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Rotational Motion
150442
A body is rolling down an inclined plane. Its translational and rotational kinetic energies are equal. The body is a :
1 Solid sphere
2 Hollow sphere
3 Solid cylinder
4 Hollow cylinder
Explanation:
D The body is rolling down an inclined plane, then its translational and rotational kinetic energies both are equal. Rotational kinetic energy \(\left(K_{R}\right)=\frac{1}{2} I \omega^{2}\) Where, moment of inertia \(=\mathrm{I}\) Angular velocity \(=\omega\) Translational kinetic energy \(\left(\mathrm{K}_{\mathrm{T}}\right)=\frac{1}{2} \mathrm{mv}^{2}\) \(\mathrm{K}_{\mathrm{T}}=\frac{1}{2} \mathrm{~m}(\mathrm{r} \omega)^{2} \quad \ldots .(\mathrm{i}) \quad(\because \mathrm{v}=\mathrm{r} \omega\) Solid sphere - moment of inertia of solid sphere \(\mathrm{I}=\frac{2}{5} \mathrm{mr}^{2}\) \(\therefore \quad \mathrm{K}_{\mathrm{R}}=\frac{1}{2} \mathrm{I} \omega^{2}\) \(\mathrm{K}_{\mathrm{R}}=\frac{1}{2} \cdot \frac{2}{5} \mathrm{mr}^{2} \cdot \omega^{2}=\frac{1}{5} \mathrm{mr}^{2} \omega^{2}\) So, \(\mathrm{K}_{\mathrm{R}} \neq \mathrm{K}_{\mathrm{T}}\) From eqn. (i) Solid cylinder - moment of inertia of solid cylinder \(\mathrm{I}=\frac{1}{2} \mathrm{mr}^{2}\) \(\mathrm{K}_{\mathrm{R}}=\frac{1}{2}\left(\frac{1}{2} \mathrm{mr}^{2}\right) \omega^{2}\) \(\mathrm{K}_{\mathrm{R}} =\frac{1}{4} \mathrm{mr}^{2} \omega^{2}\) \(\mathrm{~K}_{\mathrm{R}} \neq \mathrm{K}_{\mathrm{T}}\) Hollow cylinder- moment of inertia of hollow cylinder is \(\therefore \quad \mathrm{I} =\mathrm{mr}^{2}\) \(\quad \mathrm{~K}_{\mathrm{R}} =\frac{1}{2} \mathrm{I} \omega^{2}\) \(\mathrm{~K}_{\mathrm{R}} =\frac{1}{2} \mathrm{mr}^{2} \omega^{2}\) So, \(\mathrm{K}_{\mathrm{R}}=\mathrm{K}_{\mathrm{T}}\) So, option (d) is correct answer.
UPSEE - 2004
Rotational Motion
150444
One solid sphere and disc of same radius are falling along an inclined plane without slipping. One reaches earlier than the other due to
1 different radius of gyration
2 different sizes
3 different friction
4 different moment of inertia
Explanation:
A (i) For solid sphere, the moment of inertia about the diameter is \(\mathrm{I}_{\mathrm{s}}=\frac{2}{5} \mathrm{MR}^{2}\) Now \(\mathrm{I}=\mathrm{MK}^{2}\) for anybody, where \(\mathrm{K}\) is radius of gyration of that body. So \(\mathrm{MK}^{2}=\frac{2}{5} \mathrm{MR}^{2} \Rightarrow \mathrm{K}=\mathrm{R} \sqrt{2 / 5}\) (ii) The moment of inertia of disc about an axis passing through its centre \ perpendicular to plane is \(I_{d} =\frac{M R^{2}}{2}=M K^{2}\) \(K =R \sqrt{\frac{1}{2}}\) Now acceleration of anybody which is rolling on an inclined plane is \(a=\left(\frac{g \sin \theta}{1+K^{2} / R^{2}}\right)\) For same R, \(\mathrm{a} \propto \frac{1}{\mathrm{~K}^{2}}\) \(\Rightarrow \quad \mathrm{a}_{\mathrm{s}}>\mathrm{a}_{\mathrm{d}}\) So, solid sphere will reach earlier to bottom of an inclined plane than disc.
BITSAT-2008
Rotational Motion
150445
The ratio of the accelerations for a solid sphere (mass ' \(m\) ' and radius ' \(R\) ') rolling down an incline of angle ' \(\theta\) ' without slipping and slipping down the incline without rolling is :
1 \(5: 7\)
2 \(2: 3\)
3 \(2: 5\)
4 \(7: 5\)
Explanation:
A For solid sphere rolling without slipping on inclined plane, acceleration \(a_{1}=\frac{g \sin \theta}{\left(1+\frac{K^{2}}{R^{2}}\right)}\) For solid sphere slipping on inclined plane without rolling, acceleration \(a_{2}=g \sin \theta\) For solid sphere, \(\mathrm{I}=\frac{2}{5} \mathrm{MR}^{2}\) \(\mathrm{MK}^{2} =\frac{2}{5} \mathrm{MR}^{2}\) \(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}} =\frac{2}{5}\) Therefore, required ratio, \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{1}{\left(1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)}=\frac{1}{\left(1+\frac{2}{5}\right)}=\frac{5}{7}\) \(\text { i.e. } \quad a_{1}: a_{2}=5: 7\)
AIPMT-2014
Rotational Motion
150446
A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in figure. At any instant, for the lower most point of the disc-
1 velocity is \(\mathrm{v}\), acceleration is zero
2 velocity is zero, acceleration is zero
3 velocity is \(v\), acceleration is \(v^{2} / R\)
4 velocity is zero, acceleration is \(v^{2} / R\)
Explanation:
D Velocity at point A is 0 (zero) because disc performing pure rolling. Acceleration at point \(\mathrm{A}\) is, We know that \(\begin{aligned} \mathrm{v} =\mathrm{R} \omega \\ \text { Or } \quad \omega =\frac{\mathrm{v}}{\mathrm{R}}\end{aligned}\) Angular acceleration is \(\alpha)=R \omega^{2}\) \(\alpha=\mathrm{R} \cdot \frac{\mathrm{v}^{2}}{\mathrm{R}^{2}}=\frac{\mathrm{v}^{2}}{\mathrm{R}}\) So, Acceleration at lower most point of disc is \(\frac{\mathrm{v}^{2}}{\mathrm{R}}\).
