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solong [7]
2 years ago
13

Consider two cylindrical objects of the same mass and radius. Object A is a solid cylinder, whereas object B is a hollow cylinde

r. Part (a) If these objects roll without slipping down a ramp, which one will reach the bottom of the ramp first? MultipleChoice : 1) There is not enough information to determine 2) Object B 3) They will reach the bottom at the same time. 4) Object A Part (b) How fast, in meters per second, is object A moving at the end of the ramp if it's mass is 340 g, it's radius 31 cm, and the height of the beginning of the ramp is 41 cm? Numeric : A numeric value is expected and not an expression. VA= _____
Part (c) How fast, in meters per second, is object B moving at the end of the ramp if it rolls down the same ramp? Numeric : A numeric value is expected and not an expression. VB= ______
Physics
1 answer:
sergiy2304 [10]2 years ago
3 0

The thing whose moment of inertia is lowest will sink to the bottom first.

<h3>What distinguishes a solid from a hollow cylinder?</h3>
  • All of the mass in the hollow cylinder, ring, or hoop is located r distances from the axis of rotation.
  • However, the mass is evenly distributed throughout the solid cylinder.
  • This indicates that the solid cylinder's "rotational mass" is less.
  • A cylinder is a solid surface created by a line traveling perpendicular to a fixed line, with the end of the line describing a closed figure in a plane.
  • When initially at rest, two cylindrical objects with the same mass and radius, one solid and one hollow, roll down the same inclined plane without slipping.

E=mgh+mv2/2+Iw2/2\\v=rw\\E=mgh+mv2/2+Iv2/2r2\\\\Σmr2

To learn more about a solid from a hollow cylinder refer to:

brainly.com/question/18648373

#SPJ4

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A horizontal circular platform (m = 119.1 kg, r = 3.23m) rotates about a frictionless vertical axle. A student (m = 54.3kg) walk
Murrr4er [49]

Answer:

\omega_2=5.1rad/s

Explanation:

Since there is no friction angular momentum is conserved. The formula for angular momentum thet will be useful in this case is L=I\omega. If we call 1 the situation when the student is at the rim and 2 the situation when the student is at r_2=1.39m from the center, then we have:

L_1=L_2

Or:

I_1\omega_1=I_2\omega_2

And we want to calculate:

\omega_2=\frac{I_1\omega_1}{I_2}

The total moment of inertia will be the sum of the moment of intertia of the disk of mass m_D=119.1 kg and radius r_D=3.23m, which is I_D=\frac{m_Dr_D^2}{2}, and the moment of intertia of the student of mass m_S=54.3kg at position r (which will be r_1=r=3.23m or r_2=1.39m) will be I_{S}=m_Sr_S^2, so we will have:

\omega_2=\frac{(I_D+I_{S1})\omega_1}{(I_D+I_{S2})}

or:

\omega_2=\frac{(\frac{m_Dr_D^2}{2}+m_Sr_{S1}^2)\omega_1}{(\frac{m_Dr_D^2}{2}+m_Sr_{S2}^2)}

which for our values is:

\omega_2=\frac{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(3.23m)^2)(3.1rad/s)}{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(1.39m)^2)}=5.1rad/s

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3 years ago
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A ball is thrown straight up into the air with an initial velocity of 50 ft/sec. The height h(t) of the ball after t seconds is
Scrat [10]

Answer:

The average velocity for the time period beginning when t=1 and lasting 0.1 seconds = 16.40 ft/s.

Explanation:

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\rm h(t) = (50 t-16t^2)\ ft.

The average velocity of an object is defined as the total displacement covered by the particle divided by the total time taken in covering that displacement.

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Thus, the average velocity of the ball for the time interval \rm t_2-t_1  is given by

\rm v_{av}=\dfrac{h_2-h_1}{t_2-t_1}.

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