The flying cow problem. Find the period of rotation of the cow.

The rotating cow is attached to the pivot by a string, thus its motion requires a definite period. The period of rotation of the cow is 1.85 s.

The period of rotation of an object is the time that would be required for the object to complete one cycle or revolution.

Thus from the given diagram, let the length of the string holding the cow to the pivot be represented by x. So that applying the Pythagoras theorem, we have:

= +

= +

   = 7225

x =

  = 85

Thus , the length of the string is 85 cm.

So that, the period of rotation of the cow can be determined as follows:

Period, T = 2

where l is the length of the string and g is the gravitational force.

l = x = 85 cm (0.85 m), g = 9.8 m/

Therefore,

T = 2 x x

  = x 0.29451

  = 1.85121

T = 1.85 s

See more about Period of rotation at: link

1.26 s

Explanation:

Draw a free body diagram.  There are three forces: weight force mg pulling down, vertical tension component Tᵧ pulling up, and horizontal tension component Tₓ pulling radially.

Sum of forces in the vertical direction:

∑F = ma

Tᵧ − mg = 0

Tᵧ = mg

Sum of forces in the centripetal direction:

∑F = ma

Tₓ = m v² / r

Tension force must be in the same direction as the string, so:

Tₓ / Tᵧ = 75 / 40

40 Tₓ = 75 Tᵧ

40 m v² / r = 75 mg

8 v² / r = 15 g

v = √(15gr / 8)

Plugging in values:

v = √(15 × 10 m/s² × 0.75 m / 8)

v = 3.75 m/s

The circumference of the path is:

C = 2πr

C = 2π (0.75 m)

C = 4.71 m

So the period is:

t = C / v

t = 4.71 m / 3.75 m/s

t = 1.26 s

Round as needed.

i don't understand. explain it clearly

i had this question and the answer was 3.15 miles.

martina travels from school to her mom's office. from there, she walks to school. after school, she walks to the library and then to a friends house before finally ending back at home. find martina's total distance traveled.

a) 0 miles  

b) 0.4 miles  

c) 1 mile  

d) 3.15 miles