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Alina [70]
3 years ago
9

A ring, cylinder, solid sphere, and hollow sphere are all released from rest from the same height on an inclined surface, at the

same time. All four objects have uniform density, have the same mass, same radius, and roll without slipping.1. Which one has the largest total kinetic energy at the bottom?a) ringb) cylinderc) solid sphered) hollow spheree) they all reach the bottom with the same total kinetic energy2. Which one reaches the bottom last?a) ringb) cylinderc) hollow sphered) solid spheree) they all reach the bottom at the same time
Physics
1 answer:
crimeas [40]3 years ago
7 0

Answer:

Explanation:

  • The expression for acceleration of the rolling body on an inclined plane is given as a = gsinФ/1 + k²/R²
  • where Ф is the angle of inclination, R is the radius, k is the radius of gyration.
  • The potential energy of the system is given as ; PE = mgh
  • The potential energy will be constant for ring, cylinder, solid sphere, and hollow sphere.
  • The total kinetic energy of the rolling body is ; KE = mv²/2 + Iw²/2
  • Hence, the total kinetic energy of the ring, cylinder, solid sphere and hollow sphere will be constant.

2. The moment of inertia of the ring is given as ;

I = mR²

The moment of inertia of the ring is maximum and therefore reaches the bottom last.

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PLEASE HELP
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The vertical component of the initial velocity is v_0_y = \frac{y}{t} + \frac{1}{2} gt

The horizontal component of the initial velocity is v_0_x = \frac{x}{t}

The horizontal displacement when the object reaches maximum height is X = \frac{xy}{gt^2} + \frac{x}{2}

The given parameters;

the horizontal displacement of the object, = x

the vertical displacement of the object, = y

acceleration due to gravity, = g

time of motion, = t

The vertical component of the initial velocity is given as;

y = v_0_yt - \frac{1}{2} gt^2\\\\v_0_yt = y + \frac{1}{2} gt^2\\\\v_0_y = \frac{y}{t} + \frac{1}{2} gt

The horizontal component of the initial velocity is calculated as;

x = v_0_xt\\\\v_0_x = \frac{x}{t}

The time to reach to the maximum height is calculated as;

T = \frac{v_f_y -v_0_y}{-g} \\\\T = \frac{-v_0_y}{-g} \\\\T = \frac{v_0_y}{g} \\\\T =  \frac{1}{g}  (v_0_y)\\\\T = \frac{1}{g} (\frac{y}{t} + \frac{1}{2} gt)\\\\T = \frac{y}{gt} + \frac{1}{2} t

The horizontal displacement when the object reaches maximum height is calculated as;

X= v_0_x \times T\\\\X= \frac{x}{t} \times (\frac{y}{gt} + \frac{1}{2} t)\\\\X = \frac{xy}{gt^2} + \frac{x}{2}

Learn more here: brainly.com/question/20689870

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