# Physic I

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1) Work is energy transferred to an object by a force acting on the object, since energy is measured in Joules (J) work is also measured in Joules.  The formula for work, is `` .  The force on the elevator is 5500N, and its distance is 50.0m so the work done on the elevator is ``J

2) The hiker has a mass of 55.6 kg, and gravitational force on the surface of the earth is mass (in kg) x 9.8 `` so Gravity has a force of `` 544.88 N on the hiker.  If the hiker moves upward, Gravity does negative work, and that work is the force of gravity on the hiker, times his upward (negative) distance of 50 m, which is `` -27,400J

3) Since work is energy transferred to an object by a constant force, when work ceases, the object should have kinetic energy equal to the work done on it (assuming no friction acts on the object).  The work done on the cart is the force applied (200N) times its distance (10m) `` .  So the cart should have 2000J of kinetic energy.

4) Work done on the car is energy given to the car.  The car was already in motion, so it already had energy.  The formula for an object's kinetic energy is `` , filling in the car's mass (1208 kg), and velocity (10 m/s), we find that the car had `` 60,400J to begin with.  400,000J of work is done on the car, giving the car that much more kinetic energy, so the car now has `` 460,400J of Kinetic energy.  We solve the formula for v `` .  We fill in the values we have K=460,400. m=1208 and we find that the final velocity of the car is `` the final velocity is 27.61 m/s

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Physic I

The moment of inertia of a given distribution of mass with respect to an axis of rotation is by DEFINITION

`I =int R^2*dm`    (1)

where the integral is taken for all values of distance `R` measured from the axis of rotation to the infinitesimal mass `dm` . The case of a plane (x,y) distribution of mass is given in the figure below (the axis of rotation is perpendicular to the figure):

`I_z = int (x^2+y^2)*dm`

This means that the moment of inertia is maximum when the mass is distributed at higher distances from the rotation axis.

The highest moment of inertia will have the flat hoop, since for it all mass is distributed at the highest distance `R = R_(max)` from the center.

Now, between the sphere (uniform) and the upright hoop both have the mass distributed uniform from `R=0` to `R=R_max` , but the sphere has a symmetrical distribution over all three x,y,z coordinates, whereas the upright hoop has a symmetrical distribution over only two coordinates. This means that the infinitesimal mass `dm=M/V` will be larger for hoop than for the sphere and the moment of inertia of the sphere (uniform) will be the smallest (see equation (1)).

Thus the flat hoop will have the highest moment of inertia, while the sphere will have the smallest moment of inertia.

Indeed, the computations give:

For the flat hoop `I =MR^2`

` ` For the upwright hoop `I =(MR^2)/2`

For the sphere `I =(2MR^2)/5`

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Physic I

a) The length of the string for a complete rotation of the yo-yo is

`d_0 =2*pi*R =2*pi*0.005 =0.0314 m`

For a total length of the string `d = 30 cm=0.3 m` the number of revolutions the yo-yo need to make is

`N =d/d_0 =0.3/0.0314 =9.55 revolutions`

b)

The linear acceleration of the yo-yo comes from

`d =v_0*t + (a*t^2)/2`  with initial speed `v_0 =0 m/s`

Thus `a =(2*d)/t^2 =(2*0.3)/25^2=9.6*10^-4 m/s^2`

This linear acceleration is tangent to the yo-yo disk. The angular acceleration `epsilon` (perpendicular to the yo-yo disk plane) comes from

`a =epsilon*R rArr epsilon = a/R =(9.6*10^-4)/(5*10^-3) =0.192 (rad)/s^2`

(The above relation is analog `v = omega*R` )

c)

The angular velocity as a function of time equation is

`omega=omega_0 +epsilon*t`   (analog `v=v_0 +a*t` )

For `omega_0 =0 (rad)/s` we have

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Physic I

The figure is below attached.

a)

For the falling mass the free body diagram includes the weight of the mass `G =m*g (=m*a_g)` directed downwards, the string tension force `F_T` directed upwards, and the inertia of the mass itself `m*a` directed upwards (we suppose the mass is falling down).

b)

On the vertical axis the sum of all forces need to be zero.

`F_T +m*a =m*g rArr F_T =m*(g-a)`

Of course if we impose the condition `a < <g rArr F_T ~~m*g`

c)

The string is wrapped around the spool. It means that the torque on the system is simply (the angle between the force tension in the string `F_T` and the spool radius `r_s` is 90 degree):

`T =F_T*r_s`

d)

The angular acceleration `epsilon(=alpha)` and linear acceleration `a` are related by the equation:

`a =epsilon*r_d rArr epsilon =a/r_d`

(similar to `v = omega*r_d` )

e)

The equation that relates the torque `T` , the moment of inertia `I_d` and the angular acceleration `epsilon (=alpha)` is

`T =I_d*epsilon`

(This is similar to the Newton second law `F =M*a` )

f)

For a uniform disk of radius `r_d` and mass `M_d` the moment of inertia about a perpendicular axis on disk plane (on its center) is

`I_d =(M_d*r_d^2)/2`