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Physics System of Particles and Rotational Motion

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3 months ago

Physics System of Particles and Rotational Motion Class 11th

Four point masses, each of mass m, are fixed at the corners of a square of side l. The square is rotating with angular frequency ω, about an axis passing through one of the corners of the square and parallel to its diagonal, as shown in the figure. The angular momentum of the square about this axis is:

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Vishal Baghel

Contributor-Level 10

I = m (l/√2)²*2 + m (√2l)² = 3ml²
L=Iω = 3ml²ω

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Physics System of Particles and Rotational Motion Class 11th

Shown in the figure is a hollow icecream cone (it is open at the top). If its mass is M, radius of its top, R and height, H, then its moment of inertia about its axis is:

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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Physics System of Particles and Rotational Motion Class 11th

A thin rod of mass 0.9 kg and length 1 m is suspended, at rest, from one end so that in can freely oscillate in the vertical plane. A particle of move 0.1 kg moving in a straight line with velocity 80 m/s hits the rod at its bottom most point and sticks to it (see figure). The angular speed (in rad/s) of the rod immediately after the collision will be

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Vishal Baghel

Contributor-Level 10

L_i = L_f
mvL = Iω
mvL = (ML³/3 + mL²)ω

Before collision
After collision
0.1 * 80 * 1 = (0.9 * 1²)/3 + 0.1 * 1²)ω
8 = (3/10 + 1/10)ω 8 = (4/10)ω ω = 20 rad/sec

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Physics System of Particles and Rotational Motion Class 11th

Consider a uniform rod of mass $M = 4 m$ and length $l$ pivoted about its centre. A mass $m$ moving with velocity $v$ making angle $θ = \frac{π}{4}$ $\frac{}{}$ to the rod's long axis collides with one end the rod and sticks to it. The angular sped of the rod-mass system just after the collision is

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Raj Pandey

Contributor-Level 9

$L_{o i} = L_{o f}$

$\begin{array}{r} \frac{m v}{\sqrt[]{2}} * \frac{l}{2} = [\frac{4 m l^{2}}{12} + \frac{m l^{2}}{4}] * ω \\ ω = \frac{6 v}{7 \sqrt[]{2} l} = \frac{3 \sqrt[]{2} v}{7 l} \end{array}$

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Physics System of Particles and Rotational Motion System of Particles and Rotational Motion

As shown in the figure, $a$ bob of mass $m$ is tied by $a$ massless string whose other end portion is wound on a fly wheel (disc) of radius $r$ and mass $m$ . When released from rest the bob starts falling vertically. When it has covered a distance of $h$ , the angular speed of the wheel will be

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Vishal Baghel

Contributor-Level 10

$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} l ω^{2}$

$v = ω R$ (no slipping)

$m g h = \frac{1}{2} m ω^{2} R^{2} + \frac{1}{2} \frac{m R}{2} ω^{2}$

$m g h = \frac{3}{4} m ω^{2} R^{2}$

$ω = \sqrt[]{\frac{4 g h}{3 R^{2}}} = \frac{1}{R} \sqrt[]{\frac{4 g h}{3}}$

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Physics System of Particles and Rotational Motion System of Particles and Rotational Motion

Three point particles of masses $10 k g, 1.5 k g$ and $2.5 k g$ are placed at there corners of a right angle triangle of sides $4.0 c m, 3.0; c m$ and $5.0 c m$ as shown in the figure. The centre of mass of the system is at a point:

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Vishal Baghel

Contributor-Level 10

Take $1 k g$ mass at origin

$X_{c m} = \frac{1 * 0 + 1.5 * 3 + 2.5 * 0}{5} = 0.9 c m$

$Y_{c m} = \frac{1 * 0 + 1.5 * 0 + 2.5 * 4}{5} = 2 c m$

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Physics System of Particles and Rotational Motion Class 11th

The radius of gyration of a uniform rod of length $l$ , about an axis passing through a point $\frac{l}{4}$ $\frac{}{}$ away from the centre of the rod, an perpendicular to it is:

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Vishal Baghel

Contributor-Level 10

$\frac{M l^{2}}{12} + M (\frac{l}{4}) = {M K}^{2}$

$\frac{l^{2}}{12} + \frac{l^{2}}{16} = K^{2}$

$K = \sqrt[]{\frac{7}{48}} l$

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Physics System of Particles and Rotational Motion Class 11th

The coordinates of centre of mass of a uniform flag shaped lamina (thin flat plate) of mass 4 $k g$ . (the coordinates of the same are shown in figure) are:

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Raj Pandey

Contributor-Level 9

$(0,3)$

$\begin{array}{r} {\overset{?}{r}}_{c m} = \frac{1 * (\frac{\hat{i}}{2} + \hat{j}) + 1 * (\hat{i} + \frac{5 \hat{j}}{2})}{2} \\ {\overset{?}{r}}_{c m} = \frac{3}{4} \hat{i} + \frac{7}{4} \hat{j} \end{array}$

$(0,0)$

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Physics System of Particles and Rotational Motion Class 11th

Mass per unit are of a circular disc of radius a depends on the distance $r$ from its centre as $σ (r) = A + B r$ . The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its centre is:

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alok kumar singh

Contributor-Level 10

Moment of inertia of ring

$d l = {d m r}^{2}$

$I = \int_{0}^{a} ? (A + B r) 2 π r d r π r^{2}$

$= 2 π A \int_{0}^{a} ? r^{3} d r + 2 π B \int_{0}^{a} ? r^{4} d r$

$= 2 π [A \frac{a^{4}}{4} + B \frac{a^{5}}{5}]$

= 2 π a^{4} [\frac{A}{4} + \frac{B a}{5}]

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Physics System of Particles and Rotational Motion Class 11th

Consider a 20 kg uniform circular disk of radius 0.2 m. It is pin supported at its centre and is at rest initially. The disk is acted upon by a constant force F = 20 N through a massless string wrapped around its periphery as shown in the figure. Suppose the disk makes n number of revolutions to attain an angular speed of 50 rad s-1. The value of n, to the nearest integer, is ---------.

[Given : In one complete revolution, the disk rotates by 6.28 rad ]