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zavuch27 [327]
3 years ago
13

A lawn roller in the form of a solid uniform cylinder is being pulled horizontally by a horizontal force B applied to an axle th

rough the center of the roller. The roller radius 0.59 m and mass 48 kg. What magnitude B of the force is required to produce and acceleration a=4.2 m/s^2 of the center of mass of the roller if the lawn rolls without slipping?

Physics
1 answer:
horsena [70]3 years ago
4 0

Answer:

Explanation:

mass, m = 48 kg

radius, r = 0.59 m

acceleration of center of mass, a = 4.2 m/s2

The free body diagram of cylinder is as follows:

Since the cylinder is rolling without slipping, therefore,a=αr,

which gives, angular acceleration, α = a/r = (4.2/0.59)rad/s2 = 7.12 rad/s2

Moment of inertia of cylinder about center of mass C, IC= (1/2)mr2

So, by parallel axis theorem, moment of inertia about point A,IA = (1/2)mr2 + mr2 =(3/2)mr2 = (3/2)*48*0.592 kg.m2 =25.06 kg.m2

Now, taking the moment of forces about point A, T =IAα

which gives, B*r = IAα, and we get, B =(25.06*7.12)/0.59 N = 302.42 N

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Answer:

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Explanation:

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initial temperature of water, T_i=18^{\circ}C

time taken to vapourize half a liter of water, t=18\ min=1080\ s

desity of water, \rho=1\ kg.L^{-1}

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enthalpy of vaporization of water, h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}

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Amount of heat required to raise the temperature of given water mass to 100°C:

Q_s=m.c.\Delta T

Q_s=1\times 4180\times (100-18)

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Q_v=m'\times h_{fg}

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m'=0.5\ kg= mass of water vaporized due to boiling

Q_v=0.5\times 2256.4

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