Angular momentum of the particle rotating with a central force is constant due to
Constant Force
Constant linear momentum
Zero Torque
Zero Torque
A force of acts on O, the origin of the coordinate system. The torque about the point (1, −1) is
A thin circular ring of mass m and radius R is rotating about its axis with a constant angular velocity ω. Two objects each of mass M are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity ω′
The rms value of the electric field of the light coming from the Sun is 720 N/C. The average total energy density of the electromagnetic wave is
3.3 × 10−3 J/m3
4.58 × 10−6 J/m3
6.37 × 10−9 J/m3
6.37 × 10−9 J/m3
Four point masses, each of value m, are placed at the corners of a square ABCD of side A. The moment of inertia through A and parallel to BD is
m
2m
√m
√m
A ‘T’ shaped object with dimensions shown in the figure, is lying on a smooth floor. A force F is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C
2
3
4
4
C.
4
The point P must be the centre of mass of the T-shaped object since the force F does not produce any rotational motion of the object. So, we have to find the distance of the centre of mass from the point C.
The horizontal part of the T-shaped object has length L. If the mass of the horizontal portion is ‘m’, the mass of the vertical portion of the T- shaped object is 2m since its length is 2L. For finding the centre of mass of the T shaped object, it is enough to consider two point masses m and 2m located respectively at the midpoints of the horizontal and vertical portions of the T.
Therefore, the T-shaped object reduces to two point masses m and 2m at distances 2L and L respectively from the point C. The distance ‘r’ of the centre of mass of the system from the point C is given by
r = (m1r1 + m2r2)/(m1 + m2) = (m×2L + 2m×L)/(m + 2m) = 4L/3
[ Note that we have used the equation, r = (m1r1 + m2r2)/(m1 + m2) for the position vector r of the centre of mass in terms of the position vectors r1 and r2 of the point masses m1 and m2. We could use the simple equation involving the distances from C since the points are collinear].
A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then:
A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012/sec. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avagadro number = 6.02 × 1023 gm mole–1)
5.5 N/m
6.4 N/m
7.1 N/m
2.2 N/m
A wheel rotating at 12 rev/sis brought to rest in 6 s. The average angular deceleration in rad/s2 of the wheel during this process is
−
4
72