Physics, 03.12.2019 18:31 KP1998

Consider a father pushing a child on a playground merry-go-round. the system has a moment of inertia of 84.4 kg · m2. the father exerts a force on the merry-go-round perpendicular to its 1.50 m radius to achieve a torque of 375 n · m.(a) calculate the rotational kinetic energy (in j) in the merry-go-round plus child when they have an angular velocity of 29.2 rpm. j(b) using energy considerations, find the number of revolutions the father will have to push to achieve this angular velocity starting from rest. revolutions(c) again, using energy considerations, calculate the force (in n) the father must exert to stop the merry-go-round in seven revolutions. n

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Answer from: emmazhu1106

A)The rotational kinetic energy is ≈ 9.9962 J

B)The number of revolutions to achieve that angular velocity = 0.0085

C)Force to stop it = 0.15152N

Explanation:

Given :

The system has a moment of inertia of $84.4kgm^{2}$ Radius R = 1.50m

Torque exerted = 375Nm

Rotational kinetic energy :

$E = \frac{1}{2} Iw^{2}$

where,

is the rotational kinetic energy (J)

is the moment of inertia $(Kgm^{2})$

is the angular velocity $(rads^{-1})$

⇒ $I=84.4\\w=29.2rpm=\frac{29.2}{60} rps=0.4867 rads^{-1}$

⇒ $E = \frac{1}{2} Iw^{2}\\E=\frac{1}{2} *84.4*(0.4867)^{2}\\E=9.9962 J$

Thus the energy is ≈ 9.9962 J

B) Work done by the torque = The change in rotational kinetic energy

$T\alpha = \frac{1}{2} Iw^{2} - \frac{1}{2} Iw_{o}^{2}$

where ,

is the torque exerted $\alpha$ is the angle traversed w_o

is the initial angular velocity.

Thus,

$375*\alpha =\frac{1}{2} *84.4*(0.4867)^{2}-\frac{1}{2} *84.4*0\\\alpha =0.026657 rad\\=\frac{0.026657}{\pi } rev\\=0.0085 rev$

The number of revolutions = 0.0085

7 revolutions :

$=7*2\pirad\\=14\pirad$

$T\alpha = \frac{1}{2} Iw^{2} - \frac{1}{2} Iw_{o}^{2}$

$T*14\pi =9.9962\\T= 0.2273 Nm\\F=\frac{T}{R} =\frac{0.2273}{1.5} \\F=0.15152N$

Force = 0.15152N

Answer from: Quest

What sports are they giving you

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

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Consider a father pushing a child on a playground merry-go-round. the system has a moment of inertia...