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Artyom0805 [142]
3 years ago
7

A particle moves along a circle with radius R, so that the tangential component of its acceleration is constant. At t = 0 the ve

locity of the particle was equal to zero. Find
(a) the normal component an of the acceleration as function of time.
(b) the magnitude of the acceleration vector a as well as the angle the vector a forms with the position vector r as functions of time.
Physics
1 answer:
Gala2k [10]3 years ago
7 0
Answer: a


Explanation:
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A 0.37-kg object connected to a light spring with a force constant of 23.2 N/m oscillates on a frictionless horizontal surface.
mars1129 [50]

Answer:

a) v = 31.67 cm / s , b)   v = -29.36 cm / s , c) v= 29.36 cm/s, d) x = 3.46 cm

Explanation:

The angular velocity in a simple harmonic movement is

       w = √ K / m

       w = √ 23.2 / 0.37

       w = 7,918 rad / s

a) the expression against the movement is

        x = A cos (wt + Ф)

Speed ​​is

        v = dx / dt = - A w sin (wt + Ф)

 The maximum speed occurs for cos = ± 1

        v = A w

        v = 4.0 7,918

        v = 31.67 cm / s

b) as the object is released from rest

        0 = -A w sin (0+ Фi)

        sin Ф = 0

         Ф = 0

The equation is

        x = 4.0 cos (7,918 t)

        v = -4.0 7,918 sin (7,918 t)

        v = - 31.67 sin (7.918t)

     

Let's look for the time for a displacement of x = 1.5 cm, remember that the angles must be in radians

          7,918 t = cos⁻¹ 1.5 / 4.0

          t = 1,186 / 7,918

          t = 0.1498 s

We look for speed

         v = -31.67 sin (7,918 0.1498)

         v = -29.36 cm / s

c) if the object passes the equilibrium equilibrium position again at this point the velocity has the same module, but the opposite sign

         v = 29.36 cm / s

d) let's look for the time for the condition v = v_max / 2

         31.67 / 2 = 31.67 sin ( 7,918 t)

          7.918t = sin⁻¹ 0.5

         t = 0.5236 / 7.918

         t = 0.06613

With this time let's look for displacement

         x = 4.0 cos (7,918 0.06613)

        x = 3.46 cm

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