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System of Particles and Rotational Motion

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2 months ago

System of Particles and Rotational Motion Class 11th

A metre scale is balanced on a knife edge at its centre. When two coins, each of mass 10 g are put one on the top of the other at the 10.0 cm mark the scale is found to be balanced at 40.0 cm mark. The mass of the metre scale is found to be x * 10^-2 kg. The value of x is _________.

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Vishal Baghel

Contributor-Level 10

$τ_{A} = 0$

2mg *30 = Mg *10

$\Rightarrow 2 * 0.01 g * 30 = M g * 10$

M = 6 * 10^-2kg

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2 months ago

System of Particles and Rotational Motion Physics NCERT Exemplar Solutions Class 12th Chapter Seven

A $\sqrt[]{34} m$ long ladder weighing 10kg leans on a frictionless wall. Its feet rest on the floor 3m away from the wall as shown in the figure. If F_f and F_w are the reaction forces of the floor and the wall, then ratio of F_w / F_f will be:

(Use g = 10 m/s²)

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alok kumar singh

Contributor-Level 10

mg = 10 * 10

= 100 of

NA = $F_{ω}$ reaction force offered by the wall

$\sqrt[]{N_{3}^{2} + f_{B}^{2}} = F_{f}$ reaction force offered by the floor.

By NLM 1

Translational fB = NA

NB = mg = 100N

Ac² + Bc² = AB²

Ac² = 34 – 9

Ac = 5m

Rotational equilibrium $τ_{B} = 0$

mg $\frac{l}{2} c o s θ = N_{A} l s i n θ$

$100 \frac{c o s θ}{2} = N_{A} s i n θ$

NA = 50 cot

Now cot = $\frac{3}{5}$

N_A= $50 * \frac{3}{5} = 30 N$

Therefore, NA = $F_{ω} = 30 N$ - (i)

$F_{f} = \sqrt[]{N_{B}^{2} + f_{B}^{2}} = \sqrt[]{10 0^{2} + 3 0^{2}} = 10 \sqrt[]{109} N$ - (2)

$\frac{f_{ω}}{F_{f}} = \frac{30}{10 \sqrt[]{109}} = \frac{3}{109}$

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2 months ago

System of Particles and Rotational Motion Physics NCERT Exemplar Solutions Class 12th Chapter Seven

A ball is spun with angular acceleration a = 6t² – 2t where t is in second and a is in rads^-¹. At t = 0, the ball has angular velocity of 10 rads^-¹ and angular position of 4 rad. The most appropriate expression for the angular position of the ball is:

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alok kumar singh

Contributor-Level 10

a = 6t² – 2t, of t 20, w = 10 rad/sec

, q = 4 rad

$α = \frac{d ω}{d t}$

$\Rightarrow \int_{10}^{ω} d ω = \int_{t - 0}^{t} α d t$

$\Rightarrow ω - 10 = \int_{0}^{t} (6 t^{2} - 2 t) d t$

$ω = \frac{d θ}{d t}$

$d θ = ω d t$

$\int_{4}^{θ} d θ = \int_{0}^{t} (2 t^{3} - t^{2} + 10) d t$

= $\frac{t^{4}}{2} - \frac{t^{3}}{3} + 10 t + 4$

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2 months ago

System of Particles and Rotational Motion physics ncert exemplar solutions class 12th chapter two

A set of 20 tuning forks is arranged in series of increasing frequencies. If each fork gives 4 beats with respect to the preceding fork and the frequency of the last fork is twice the frequency of the first, then the frequency of last fork is ___________Hz.

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Vishal Baghel

Contributor-Level 10

A/c to question

f + 76 = 2f

f = 76

last frequency

= f + 76 = 2 * 76

= 152 H₂

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2 months ago

System of Particles and Rotational Motion physics ncert exemplar solutions class 12th chapter two

A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is-

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Vishal Baghel

Contributor-Level 10

K.ER (Rotational) = $\frac{1}{2} I_{C M} ω^{2} = \frac{1}{2} * \frac{2}{5} M R^{2} ω^{2} = \frac{M R^{2} ω^{2}}{5}$ - (1)

$K . E_{T o t a l} = K . E_{R} + K . E_{T}$

$= \frac{M R^{2} ω^{2}}{5} + \frac{1}{2} M V_{c m^{2}}$

$= \frac{7 M R^{2} ω^{2}}{10}$ - (2)

$\frac{k . E_{R}}{K . E_{T o t a l}} = \frac{1}{5} * \frac{10}{7} = \frac{2}{7}$

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2 months ago

System of Particles and Rotational Motion Physics NCERT Exemplar Solutions Class 12th Chapter Seven

An object is thrown vertically upwards. At its maximum height, which of the following quantity becomes zero?

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Payal Gupta

Contributor-Level 10

At maximum height velocity (v) = 0

So, momentum = mv = m * 0 = 0

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2 months ago

System of Particles and Rotational Motion Physics NCERT Exemplar Solutions Class 12th Chapter Seven

A spherical of 1 kg mass and radius R is rolling with angular speed $ω$ on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is $\frac{a}{3} R^{2} ω .$ The value of a will be:

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alok kumar singh

Contributor-Level 10

L₀ = RMV_cm + I_cm $ω$

= RM $(R ω) + \frac{2}{3} M R^{2} ω$

L₀ = $\frac{5}{3} M R^{2} ω$

$\frac{5}{3} * 1 R^{2} ω = \frac{5}{3} R^{2} ω = \frac{a}{3} R^{2} ω$ =

a = 5

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New answer posted

2 months ago

System of Particles and Rotational Motion Physics NCERT Exemplar Solutions Class 12th Chapter Seven

A spherical of 1 kg mass and radius R is rolling with angular speed $ω$ on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is $\frac{a}{3} R^{2} ω .$ The value of a will be

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

L₀ = RMV_cm + I_cm $ω$

= RM $(R ω) + \frac{2}{3} M R^{2} ω$

L₀ = $\frac{5}{3} M R^{2} ω$

= $\frac{5}{3} * 1 R^{2} ω = \frac{5}{3} R^{2} ω = \frac{a}{3} R^{2} ω$

a = 5

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New answer posted

2 months ago

System of Particles and Rotational Motion Physics NCERT Exemplar Solutions Class 12th Chapter Seven

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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Vishal Baghel

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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2 months ago

System of Particles and Rotational Motion Physics NCERT Exemplar Solutions Class 12th Chapter Seven

As per the given figure, two blocks each of mass 250g are connected to a spring of spring constant 2Nm^‑1. If both are given velocity $υ$ in opposite directions, then maximum elongation of the spring is:

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Vishal Baghel

Contributor-Level 10

Let m = 250 g = 0.25 kg

By Reduced mass method

$m_{r} = \frac{m_{1} m_{2}}{m_{1} + m_{2}} = \frac{m m}{m + m} = \frac{m}{2}$

By wet

$w_{S P} = Δ K . E .$

$- \frac{1}{2} k x^{2} = 0 - \frac{1}{2} (\frac{m}{2}) {(2 v)}^{2}$

$\frac{2}{2} x^{2} = 0.25 v^{2}$

$x^{2} = 0.25 v^{2}$

$x = \frac{v}{2}$

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