A wheel in uniform motion about an axis passing through its centre and perpendicular to its plane is considered to be in mechanical (translational plus rotational) equilibrium because no net external force or torque is required to sustain its motion. However, the particles that constitute the wheel do experience a centripetal the acceleration directed towards the centre. How do you reconcile this fact with the wheel being in equilibrium?
How would you set a half wheel into uniform motion about an axis passing through the centre of mass of the wheel and perpendicular to its plane? Will you require external forces to sustain the motion ?
Wheel is a rigid body. The particles of the wheel experience centripetal acceleration towards the centre. This acceleration increases due to internal elastic forces, which cancel out in pairs. In a half wheel, the distribution of mass about its centre of mass (through which axis of rotation passes) is not symmetrical. Therefore, the direction of angular momentum of the wheel does not coincide with the direction of its angular velocity. Hence, an external torque is required to sustain the motion of the half wheel.
Three particles of masses $50\, g$, $100\, g$ and $150\, g$ are placed at the vertices of an equilateral triangle of side $1\, m$ (as shown in the figure). The $(x, y)$ coordinates of the centre of mass will be
The position vector of three particles of masses $1\, kg, 2\, kg$ and $3\, kg$ are $\overrightarrow {{r_1}} = (\widehat i + 4\widehat j + \widehat k)\,m,\overrightarrow {{r_2}} = (\widehat i + \widehat j + \widehat k)\,m$ and $\overrightarrow {{r_3}} = (2\widehat i - \widehat j - 2\widehat k)\,m$ respectively. The position vector of their centre of mass is
Write the difference between centre of gravity and centre of mass of a body?
A rod of length is $3 \;m$ and its mass acting per unit length is directly proportional to distance $x$ from one of its end then its centre of gravity from that end will be at
Center of mass of a system of three particles of masses $1, 2, 3\, kg$ is at the point $(1\, m, 2\, m, 3\, m)$ and center of mass of another group of two particles of masses $2 \,kg$ and $3\, kg$ is at point $(-1 \,m, 3\, m, -2\, m)$ . Where a $5\, kg$ particle should be placed, so that center of mass of the system of all these six particles shifts to center of mass of the first system?