The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its mid - point and perpendicular to its length is Io. Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is
Io + ML2/4
Io + 2ML2
Io + ML2
Io + ML2
A.
Io + ML2/4
The theorem of parallel axis for moment of inertia
I = ICM + Mh2
A particle moves in a circle of radius 5 cm with constant speed and time period 0.2 πs. The acceleration of the particle is
25 m/s2
36 m/s2
5 m/s2
5 m/s2
C.
5 m/s2
D.
5 m/s2
A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in the figure. When a string is cut, the initial angular acceleration of the rod is,
3g / 2L
g/L
2g/L
2g/L
A.
3g / 2L
Torque on the rod = moment of weight of the rod about P
... (i)
Moment of inertia of rod about,
... (ii)
As = I
From equations (i) and (ii), we get
The instantaneous angular position of a point on a rotation wheel is given by the equation
Q(t) = 2t3 - 6t2
The torque on the wheel becomes zero at
t = 0.5 s
t = 0.25 s
t = 2s
t = 2s
D.
t = 2s