Learn combined motion on a rigid body dynamics, rolling & static friction formula with examples. Practice radius of gyration formula, rigid body motion & rigid body dynamics, rolling friction examples.
e.g. Consider a rod having angular velocity \(\omega\) as shown in figure.
\(\omega_C=\dfrac{v}{r}\)
where \(v\) = velocity of point A
\(r\)= distance between A and C
\(\omega_C=\dfrac{2\omega x}{2x}=\omega\)
\(\omega_C=\omega\)
\(\omega_B=\dfrac{v}{r}\)
where \(v\) = velocity of point B
\(r\) = distance between A and B
\(\omega_B=\dfrac{\omega x}{x}=\omega\)
\(\omega_B=\omega\)
Conclusion
The angular velocity at every point is same about every point on the rigid body.
\(\tau_{cm}=I_{cm}\alpha\)
where
\(\tau_{cm}\) = Torque about center of mass
\(I_{cm}\) = Moment of inertia about center of mass
\(\alpha\)= Angular acceleration of rigid body (we need to specify any point, as angular acceleration is same about all points)
Conclusion
Effective point of application of pseudo force is center of mass. There is no need to shift real forces to center of mass or any other point to calculate torque.
Sum of forces = \(ma_{cm}\)
A Angular velocity of A with respect to B is \(\omega\), then angular velocity of C with respect to D is also \(\omega\), where A, B, C and D all lying on a rigid body
B Torque about center of mass = \(I_{cm}\alpha_{\text{any point}}\)
C Sum of forces = \(m\,a_{\text{any point}}\)
D Sum of forces = \(m\,a_{cm}\)
Note
A \(F-f=m\,a_{cm}\)
B \(FR+fR=I_{cm}\,\alpha\)
C \(a_{cm}=\alpha R\)
D \(f=\mu N\)
\(n=1\) for ring
\(n=\dfrac{1}{2}\) for disk
\(n=\dfrac{2}{5}\) for solid sphere
\(n=\dfrac{2}{3}\) for hollow sphere
1. Rolling Constraint
so, \(a_{cm}-\alpha R=a_{ground \;surface}\)
\(a_{cm}-\alpha R=0\)
\(a_{cm}=\alpha R\)
2. FBD of center of mass of a body
From FBD
\(mg\,sin\theta-f=ma_{cm}\) ... (1)
\(N-mg\,cos\theta=0\) ... (2)
3. FBD for torque about center of mass
Torque about center of mass is only due to f as all the other forces are passing through them.
\(fR=I_{cm}\alpha\)
\(a_{cm}=\alpha R\) ... (1)
\(mg\,sin\theta-f_s=ma_{cm}\) ... (2)
\(fR=I_{cm}\alpha\) ... (3)
So, \(f_sR=mk^2\alpha\)
\(f_s=\dfrac{mk^2}{R^2}a_{cm}\) \(\left[\therefore \alpha=\dfrac{a_{cm}}{R}\;\;\; \text{from (1)}\right]\) ....(4)
From (1) and (3)
\(mg\;sin\theta -\dfrac{mk^2}{R^2}a_{cm}=ma_{cm}\)
\(\implies mg\;sin\theta =ma_{cm}\left(1+\dfrac{k^2}{R^2}\right)\)
\(\implies a_{cm}=\dfrac{g\;sin\theta}{\left(1+\dfrac{k^2}{R^2}\right)}\)
Let \(n=\dfrac{k^2}{R^2}\)
where k = radius of gyration,
R = radius of body
So, \(a_{cm}=\dfrac{g\;sin\theta}{(1+n)}\)
A Solid sphere (\(S_1\)) and hollow cylinder (\(C_4\))
B Both solid spheres (\(S_1\)and \(S_2\))
C Both hollow cylinders (\(C_3\)and \(C_4\))
D Hollow cylinder (\(C_3\)) only
Note
A Towards left
B Towards right
C Zero
D Can't predict
Note
A \(F_1+F_2=ma_{cm}\)
B \(F_1-F_2=ma_{cm}\)
C \(F_1+F_2-f_s=ma_{cm}\)
D \(F_1R+F_2R-f_sR=ma_{cm}\)
\(n=1\) for ring
\(n=\dfrac{1}{2}\) for disk
\(n=\dfrac{2}{5}\) for solid sphere
\(n=\dfrac{2}{3}\) for hollow sphere
1. Rolling Constraint
so, \(a_{cm}-\alpha R=a_{ground \;surface}\)
\(a_{cm}-\alpha R=0\)
\(a_{cm}=\alpha R\)
2. FBD of center of mass of a body
From FBD
\(mg\,sin\theta-f=ma_{cm}\) ... (1)
\(N-mg\,cos\theta=0\) ... (2)
3. FBD for torque about center of mass
\(fR=I_{cm}\alpha\)
A \(a_{cm}=\alpha R\)
B \(fR=mR^2\alpha\)
C \(mg\;sin\theta-f=ma_{cm}\)
D \(f=\mu mg\;cos\theta\)