Learn angular acceleration of a pulley and Relation between Torque and Angular Acceleration. Practice free body diagrams Atwood machine problems. Find its angular acceleration at the moment it is released.
For a rotating rigid body, the angular velocity \(\omega\) and angular acceleration \((\alpha)\) of all the particles of the body are same.
Thus, \(\omega _A = \omega_B = \omega\)
Also, \(\alpha_A = \alpha_B = \alpha\)
\(ds = r_{A/O} \,d \,\theta\).........(1)
\(\dfrac{ds}{dt} = r_{A/O} \dfrac{d\,\theta}{dt}\)
\(\Rightarrow v_{A/O} = r_{A/O} \,\omega \) .......(2)
The velocity of A is perpendicular to
\( {\vec r_{A/O}}\) as well as \(\vec \omega\)
\(\dfrac{dv_{A/O}}{dt} = r _{A/O} \dfrac{d\omega}{dt}\)
\(\Rightarrow a_{A/O} = r_{A/O} \,\alpha\)
A \(a_P = R \,\alpha\) (downward) \(a_Q = R \,\alpha\) (upward)
B \(a_P = 2R\,\alpha\) (downward) \(a_Q = R\,\alpha\) (upward)
C \(a_P = 2R\, \alpha\) (downward) \(a_Q = 2R\, \alpha\) (upward)
D \(a_P = 3R\,\alpha\) (downward) \(a_Q = 2R\,\alpha\) (upward)
\(\vec \tau_0 = \vec r × \vec F\)
\(\vec \tau_0 = \vec r × (m\,\vec a)\)
\(\vec \tau_0 = \vec r × (m\, r\, \alpha)\) [ \(\because a = r \,\alpha\)]
\(\vec \tau_0 = (m\, r^2\,) \alpha\)
\(\vec \tau_0 = (I_0) \,\alpha\) [\(\because I_0 = mr^2\)]
???\(\alpha = \dfrac{\tau_{axis}}{I_{axis}}\)
Hence, angular acceleration of a body is equal to the ratio of torque acting on the body about the axis to the moment of inertia of the body about that axis.
A \(\dfrac{3g}{2\ell}\)
B \(\dfrac{4g}{\ell}\)
C \(\dfrac{2g}{\ell}\)
D \(\dfrac{g}{\ell}\)
\(\vec \tau_0 = \vec r × \vec F\)
\(\vec \tau_0 = \vec r × (m\vec a)\)
\(\vec \tau_0 = \vec r × (m\, r\, \alpha)\) [ \(\because a = r \,\alpha\)]
\(\vec \tau_0 = (m\, r^2\,) \alpha\)
\(\vec \tau_0 = (I_0) \,\alpha\) [ \(\because I_0 = mr^2\)]
\(\alpha = \dfrac{\tau_{axis}}{I_{axis}}\)
Hence, angular acceleration of a body is equal to the ratio of torque acting on the body about the axis to the moment of inertia of the body about that axis.
A \(\dfrac{F_1 R}{MK^2}\)
B \(\dfrac{F_2 R}{MK^2}\)
C \(\dfrac{(F_1 - F_2) R}{MK^2}\)
D \(\dfrac{(F_1 + F_2) R}{MK^2}\)
\(\vec \tau_0 = \vec r × \vec F\)
\(\vec \tau_0 = \vec r × (m\vec a)\)
\(\vec \tau_0 = \vec r × (m\, r\, \alpha)\) [ \(\because a = r \,\alpha\)]
\(\vec \tau_0 = (m\, r^2\,) \alpha\)
\(\vec \tau_0 = (I_0) \,\alpha\) [\(\because I_0 = mr^2\)]
\(\alpha = \dfrac{\tau_{axis}}{I_{axis}}\)
Hence, angular acceleration of a body is equal to the ratio of torque acting on the body about the axis to the moment of inertia of the body about that axis.
A \(\dfrac{m_1 g}{\left(\dfrac{MK^2}{R^2}\right)R}\)
B \(\dfrac{(m_1-m_2) g}{\left(\dfrac{MK^2}{R^2}+m_1+m_2\right)R}\)
C \(\dfrac{m_1\,g}{R}\)
D \(\dfrac{MK^2}{g} R\)
A \(\ell g \,cos\theta\)
B \(\dfrac{\ell}{2} g \,cos\theta\)
C \(\dfrac{\left(\dfrac{\ell}{2} - \ell_1\right)g \,cos\theta}{\dfrac{\ell^2}{12}+ \left(\dfrac{\ell}{2}-\ell_1\right)^2}\)
D \(\ell\,g\)
A \(\dfrac{mg}{\left(m+ \dfrac{M}{2}\right)}\)
B \(mg\)
C \(\dfrac{M}{2}\)
D \(M\)
A \(m_1g\)
B \( \dfrac{m_1\,g}{\left(m_1+m_2 + \dfrac{M}{2}\right)R}\)
C \(m_2\,g\)
D \(M\,g \cdot R\)