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Physics System of Particles and Rotational Motion

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5 months ago

Physics System of Particles and Rotational Motion Class 11th

A 2kg steel rod of length 0.6m is clamped on a table vertically at its lower end and is free to rotate in vertical plane. The upper end is pushed so that the rod falls under gravity. Ignoring the firction due to clamping at its lower end, the speed of the free end of rod when it passes through its lowest position is________ms^-1.

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alok kumar singh

Contributor-Level 10

$l = \frac{m l^{2}}{3}$

Energy conservation Low

$m g l = \frac{1}{2} \frac{m l^{2}}{3} ω^{2} . . . . . (i)$

And speed V = $ω r = ω l$

then $V = \sqrt[]{6 g l} = \sqrt[]{6 * 10 * . 6}$

->6 m/s

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5 months ago

Physics System of Particles and Rotational Motion Class 11th

A disc of mass 1kg and radius R is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest position, its angular speed will be $\sqrt[]{\frac{x}{3 R} r a d s^{- 1}}$ where x = __________.

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Payal Gupta

Contributor-Level 10

Loss in P.E. = Gain in k.E

2 mg R = $\frac{1}{2} (\frac{1}{2} m R^{2} + m R^{2}) ω^{2}$

$ω = \sqrt[]{\frac{8 g}{3 R}} = 4 \sqrt[]{\frac{g}{2 * 3 R}}$

$x = \frac{g}{2} = 5$

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5 months ago

Physics System of Particles and Rotational Motion Class 11th

The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is $2400 g c m^{2}$ . The length of the 400 g rod is nearly:

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alok kumar singh

Contributor-Level 10

Moment of inertia of $r o d = I = \frac{m l^{2}}{12}$

$\Rightarrow 2400 = 400 \frac{l^{2}}{12}$

$\Rightarrow 72 = l^{2}$

$\Rightarrow l = \sqrt[]{72} = 8.48 cm = 8.5 cm$

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5 months ago

Physics System of Particles and Rotational Motion Class 11th

A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is $v$ in the direction shown, which one of the following options is correct ( $P$ and $Q$ are any highest and lowest points on the wheel, respectively)?

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alok kumar singh

Contributor-Level 10

In the case of pure rolling,

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5 months ago

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

The centre of a wheel rolling on a plane surface moves with a speed $υ_{0} .$ A particle on the rim of the wheel at the same level as the centre will be moving at a speed $\sqrt[]{x} υ_{0} .$ Then the value of x is__________.

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Vishal Baghel

Contributor-Level 10

$V_{P} = \sqrt[]{V_{0}^{2} + V_{0}^{2}} = \sqrt[]{2} V_{0}$

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5 months ago

Physics System of Particles and Rotational Motion Class 11th

Consider a situation in which a ring, a solid cylinder and a solid sphere roll down on the same inclined plane without slipping. Assume that they start rolling from rest and having identical diameter.

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Vishal Baghel

Contributor-Level 10

Using conservation of energy:

$m g h = \frac{1}{2} l_{c m} ω^{2} + \frac{1}{2} m v^{2} = \frac{1}{2} l_{c m} \frac{V^{2}}{r^{2}} + \frac{1}{2} m v^{2}$

$v^{2} = \frac{m g h}{\frac{1}{2} \frac{l_{c m}}{r^{2}} + m v^{2}}$

So, sphere has the greatest velocity of COM and ring has least.

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6 months ago

Physics System of Particles and Rotational Motion Class 11th

A body rotating with an angular speed of 600rpm is uniformly accelerated to 1800rpm in 10 sec. The number of rotations made in the process is__________.

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Vishal Baghel

Contributor-Level 10

$?^{2} = ?_{0}^{2} + 2 ? ? ? {(\frac{2 ?}{60} * 1800)}^{2} = {(\frac{2 ?}{60} * 600)}^{2} + 2 (\frac{2 ?}{60} * \frac{1200}{10}) ?$

$? {(60 ?)}^{2} ? {(20 ?)}^{2} = 2 (4 ?) ? ? ? = \frac{80 ? * 40 ?}{2 * 4 ?} = 400 ? ? ? r a d$

Number of rotations = $\frac{?}{2 ?} = \frac{400 ?}{2 ?} = 200$

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6 months ago

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

A circular disc reaches from top to bottom of an inclined plane of length 'L' when it slips down the plane, it takes time 't₁'. When it rolls down the plane, it takes time t₂. The value of $\frac{t_{1}}{t_{2}} i s \sqrt[]{\frac{3}{x}}$ The value of x will be__________.

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alok kumar singh

Contributor-Level 10

Case – I : When disk slides down

$t_{1} = \sqrt[]{\frac{2 L}{g s i n θ}} . . . . . . . . . . (i)$

Case – II : When disk rolls down

$t_{2} = \sqrt[]{\frac{2 L}{\frac{g s i n θ}{1 + β^{2}}}} = \sqrt[]{\frac{2 L}{\frac{g s i n θ}{1 + {(\frac{k}{R})}^{2}}}} = \sqrt[]{\frac{2 L}{\frac{g s i n θ}{1 + {(\frac{R / \sqrt[]{2}}{R})}^{2}}}} = \sqrt[]{\frac{2 L}{\frac{g s i n θ}{1 + \frac{1}{2}}}} = \sqrt[]{\frac{4 L}{3 g s i n θ}} . . . . . . (i i)$

$\Rightarrow \frac{t_{1}}{t_{2}} = \sqrt[]{\frac{2 L}{g s i n θ}} * \sqrt[]{\frac{3 g s i n θ}{4 L}} = \sqrt[]{\frac{3}{2}}$

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6 months ago

Physics System of Particles and Rotational Motion Class 11th

A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass 'm' travelling along the surface hits at one end of the rod with a velocity 'u' in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses $(\frac{m}{M}) i s \frac{1}{x} .$ The value of 'x' will be__________.

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alok kumar singh

Contributor-Level 10

Using conservation of linear momentum, we can write

$m u = M v_{1} \Rightarrow v_{1} = \frac{m u}{M} . . . . . . . . (1)$

Using conservation of angular momentum about centre of mass of Rod, we can write

$v_{1} + \frac{l ω}{2} = u \Rightarrow \frac{m u}{M} + \frac{3 m u}{M} = u \Rightarrow \frac{m}{M} = \frac{1}{4}$ . (2)

Using definition of e, we can write

$v_{1} + \frac{l ω}{2} = u \Rightarrow \frac{m u}{M} + \frac{3 m u}{M} = u \Rightarrow \frac{m}{M} = \frac{1}{4}$

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New answer posted

6 months ago

Physics System of Particles and Rotational Motion Class 11th

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alok kumar singh

Contributor-Level 10

$l = \frac{m l^{2}}{3}$

Energy conservation Low

$m g l = \frac{1}{2} \frac{m l^{2}}{3} ω^{2} . . . . . (i)$

And speed V = $ω r = ω l$

then $V = \sqrt[]{6 g l} = \sqrt[]{6 * 10 * . 6}$

->6 m/s

0 0

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