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Arada [10]
3 years ago
9

A bowling ball weighing 71.7 N is attached to the ceiling by a rope of length 3.73 m . The ball is pulled to one side and releas

ed; it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is 4.60 m/s. At this instant, what are:
a. the acceleration of the bowling ball, in magnitude and direction
b. the tension in the rope?
Physics
1 answer:
vekshin13 years ago
4 0

Answer:

A) a = 5.673 m/s²

The direction will be upwards vertically towards the point where it is suspended.

B) T = 113.2 N

Explanation:

A) We are given;

Weight of bowling ball; W = 71.7 N

Speed; v = 4.6 m/s

Rope length; r = 3.73 m

Now, formula for the centripetal acceleration is;

a = v²/r

Thus; a = 4.6²/3.73

a = 5.673 m/s²

The direction will be upwards vertically towards the point where it is suspended.

B) since weight is 71.7 N, it means that;

Mass = weight/acceleration = 71.7/9.8

Mass(m) = 7.316 kg

Thus,

Centripetal force is;

F_cent = 7.316 × 5.673

F_cent = 41.5 N

Thus, Tension in the rope is;

T = W + F_cent

T = 71.7 + 41.5

T = 113.2 N

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A circular disc of mass 20kg and radius 15cm is mounted in an horizontal cylindrical axle of radius
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Using the concepts of energy, rotational Newton's second law and rotational kinematics we can find the kinematic energy of the system formed by the disk and the cylindrical axis

          KE = 0.23 J

given parameters

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  • Cylinder radius r = 1.5 cm = 0.0015 m
  • Disk mass M = 20 kg
  • Time t = 1.2 s
  • Force F = 12 N

to find

  • Kinetic energy (KE)

This exercise must be solved in parts:

1st part. Endowment kinetic energy is the energy due to the circular motion of an object and is described by the equation

         KE = ½ I w²

Where KE is the kinetic energy, I the moment of inertia and w the angular velocity

The moment of inertia is a magnitude that measures the inertia for rotational movement, it is a scalar quantity, therefore it is additive. In this system it is composed of two bodies, the disk and the cylindrical axis, for which the total moment of inertia it is

         I_{ total} = I_{ disk} + I_{ cylinder}

the moments of inertia with respect to an axis passing through the center of mass are tabulated

disk          I_{disk} = ½ M R²

cylinder   I_{cylinder} = ½ m r²

where M and m are the masses of the disk and cylinder respectively, R and r their radii

         I_{total} = ½ (M R² + m r²) = ½ M R² ( 1 + \frac{m}{M} \ (\frac{r}{R})^2 )

         I_{total} = ½ M R² ( 1+ \frac{m}{20}  (\frac{0.015}{0.15} )^2 ) = \frac{1}{2} M R² (1 + 0.005 m)

As the shaft mass  is much lighter than the disk mass , the last term is very small, which is why we despise it.

         I_{total} = ½ M R²

2nd part. Let's use Newton's second law for endowment motion

        τ = I α

        α = \frac{\tau }{I_{total}}l

        τ = F R

        α = \frac{F \ R}{I_{total}}

With the rotational kinematics expressions, we assume that the system starts from rest (w₀ = 0)

        w = w₀ + α  t

where w is the angular velocity, alpha is the angular acceleration and t is the time

        w = 0 + \frac{\tau }{I_{total}} \ t

we substitute in the kinetic energy equation

        KE = ½ I_{total}  ( \frac{ \tau }{I_{total}} \ t )²

        KE = ½ \frac{ \tau^2 }{I_{total}} \ t^2

let's substitute

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        KE = F² R² t² / M

let's calculate

        KE = 12² 0.15² 1.2² / 20

        KE = 0.23 J

With the concepts of energy and rotational kinematics we can find the kinetic energy of the system is

       KE = 0.23 j

learn more about rotational kinetic energy here:

brainly.com/question/20261989

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