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1. A ball with a diameter of ๐
is rolling purely on a horizontal surface. Its initial velocity is ๐ฃ 0 = 1 ๐๐ . The line of action (tractrix) of the equivalent substitution of forces exerted by the ground crosses the plane of the ground in front of the projection of the center of mass
with the distance ๐ฟ. What is the value of
๐
if the ball stops^ in^ ๐ก^0 = 5๐ ^ (supposing constant deceleration)? What is the minimum value of the coefficient of friction? (The point is that the ground is not totally rigid.)
Solution:
The equation of motion in components: ๐๐๐ฅ = โ๐น๐ ๐๐๐ฆ = ๐น๐ โ ๐๐ ๐๐ฆ = 0 โ ๐น๐ = ๐๐ The equation for the rotation around the axis crossing the center of mass and perpendicular to the ๐ฅ๐ฆ-plane: ฮ๐ถ(๐ง)๐ฝ๐ง = ๐น๐๐
โ๐น๐๐ฟ. For a ball:
ฮ๐ถ(๐ง)^ =
The condition of the pure rolling is ๐ฃ = ๐
๐ โ ๐๐ฅ = ๐
๐ฝ๐ง โ ๐ฝ๐ง = ๐ ๐
๐ฅ. With these relations 2 5 ๐๐
5 ๐๐๐ฅ^ = ๐น๐โ๐น๐
๐
โ^
5 ๐น๐^ = ๐น๐
๐น๐^ =
For the static frictional force: ๐น๐ โค ๐๐น๐ = ๐๐๐ โ ๐|๐๐ฅ| โค ๐๐๐ โ |๐๐๐ฅ |โค ๐ โ
0.02 โค ๐.
2. In the arrangement showing in the figure the mass of the crab is ๐, that of the small body is ๐, the radii of the cylinders are ๐ and ๐
. The motion of the body with initial velocity ๐ฃ 0 = 0 starts at the distance โ from the ground. At this moment the distance of the center of mass of the crab from the ground is โโฒ. What will be the velocity of the body when reaches the ground? Solve the problem using a, the equations of motion, b, the principle (theorem) of work (work equation), c, the principle (theorem) of energy.
Solution:
a, The equations of motion: ๐๐ = ๐๐ โ ๐พ ๐๐๐ถ = ๐๐ + ๐พโฒ^ โ ๐น The equation for the rotation around the axis crossing the center of mass and perpendicular to the ๐ฅ๐ฆ-plane: ฮ๐ถ(๐ง)๐ฝ๐ง = ๐น๐
+ ๐พโฒ๐
Because of Newtonโs III. law ๐พ = ๐พโฒ.
Constraint relations: ๐๐ถ = ๐
๐ฝ๐ง
๐ = ๐๐ถ + ๐๐ฝ๐ง
From these equations we can calculate the value of ๐,
then using the value โ we can get the wanted ๐ฃ.
and when the small body touches the ground then
๐ธ 2 =
2 +^1
2 +^1
๐ง^2 + ๐๐โโฒโฒ.
From the conservation of the mechanical energy ๐ธ 1 = ๐ธ 2 , that is
๐๐โ + ๐๐โโฒ^ =
2 +^1
2 +^1
๐ง^2 + ๐๐โโฒโฒ.
The constraint relations are the same as in the case b,.
With them 1 2 ๐๐ฃ
2 +^1
2 +^1
๐ง^2 = ๐๐โ + ๐๐(โโฒ^ โ โโฒโฒ)
2 +^1
2 +^1
๐ง^2 = ๐๐โ + ๐๐๐
^
exactly the same equation which we have got in b,.
Calculations: โฆ
2 ๐
+ ๐ +
๐๐^2 + ๐๐
^2
3. A 1.4๐๐ bucket of a well is attached to an axle by a (weightless) rope. We drop the bucket into the 10๐ deep well. What is the velocity of the bucket when it reaches the bottom of the well, if the axle is a homogeneous cylinder with a mass of 7๐๐?
Solution:
๐ = 1.4๐๐ โ = 10๐ ๐ = 7๐๐ ๐ฉ๐ถ(๐ง)^ =
2
From the conservation of the mechanical energy
2 +^1
๐ง^2
Constraint relation: ๐ฃ = ๐
๐๐ง
2 +^1
2 ๐
^2 ๐๐โ =
2 +^1
2
