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A rolling wheel of 12kg is on an inclined plane at position P and connected to a mass of 3kg through a string of fixed length and pulley as shown in figure.

Consider PR as friction free surface. The velocity of centre of mass of the wheel when it reaches at the bottom Q of the inclined plane PQ will be  1 2 x g h   m / s .  The value of x is…………

0 18 Views | Posted 3 months ago
Asked by Shiksha User

  • 1 Answer

  • P

    Answered by

    Payal Gupta | Contributor-Level 10

    3 months ago

    By conservation of Energy

    m g h = 1 2 l P ω 2                

    3gh = 1 2 ( M R 2 + M R 2 ) ω 2  

    3 g h = m v 2                

    v 2 = 3 g h 1 2                

    v = g h 2 = x g h 2          

               

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A 3 4   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 Ff and Fw are the reaction forces of the floor and the wall, then ratio of Fw / Ff will be:

(Use g = 10 m/s2)

 
A
alok kumar singh

mg = 10 × 10

= 100 of

NA = Fω reaction force offered by the wall

N32+fB2=Ff 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 τB=0

mg l2cosθ=NAlsinθ

100cosθ2=NAsinθ

NA = 50 cot

Now cot = 35

NA = 50×35=30N

Therefore, NA = Fω=30N - (i)

Ff=NB2+fB2=1002+302=10109N - (2)

fωFf=3010109=3109

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:
A
alok kumar singh

a = 6t2 – 2t,        of t 20, w = 10 rad/sec

, q = 4 rad

α=dωdt

⇒∫10ωdω=∫t−0tαdt

⇒ω−10=∫0t (6t2−2t)dt

ω=dθdt

dθ=ω dt

∫4θdθ=∫0t (2t3−t2+10)dt

= t42−t33+10t+4

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.
V
Vishal Baghel

A/c to question

f + 76 = 2f

f = 76

last frequency

= f + 76 = 2 × 76

= 152 H2

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-
V
Vishal Baghel

K.ER (Rotational) = 12ICMω2=12×25MR2ω2=MR2ω25 - (1)

K.ETotal=K.ER+K.ET

=MR2ω25+12MVcm2

=7MR2ω210 - (2)

k.ERK.ETotal=15×107=27

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

At maximum height velocity (v) = 0

So, momentum = mv = m × 0 = 0

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