Physics, 09.01.2020 05:31 cathydaves

Ahoop and a solid disc are relased from rest
atthe top of an incline and allowed to roll down the incline
withoutslipping.
which is correct?
estimate the angular acceleration of a
smallpebble stuck to a bicyle tire as the bicycle accelerates from
restto 10 mph (4.47 m/s) in 2 seconds.

Answers: 2

Show answers

Answers

Answer from: alexscott

1) The hoop and a solid disc rolling without slipping down an incline plane.

Their final velocities are proportional to their moment of inertia.

The condition for moment of inertia: v = ωR

We will use conservation of energy.

For the hoop:

$K_1 + U_1 = K_2 + U_2\\0 + m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}I\omega_h^2 + 0$

They are released from rest, so their initial kinetic energy is zero. And when they reach the bottom, their final potential energy is also zero.

The moment of inertia of a hoop is

I_h = m_hR^2

Let's continue with the energy equations:

$m_h gh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}(m_hR^2)(\frac{v_h^2}{R^2})\\m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}m_hv_h^2\\m_hgh = m_hv_h^2\\v_h = \sqrt{gh}$

Similarly for the solid disk with a moment of inertia of (1/2)mR^2:

$K_1 + U_1 = K_2 + U_2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}I_d\omega_d^2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}(\frac{1}{2}m_dR^2)(\frac{v_d^2}{R^2})\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{4}m_dv_d^2\\m_dgh = \frac{3}{4}m_dv_d^2\\v_d = \sqrt{\frac{4gh}{3}}$

Comparing the final velocities, we can conclude that the solid disk reaches the bottom first.

2) The angular acceleration of the pebble is equal to the angular acceleration of the tire, since they stuck together. We can deduce the angular acceleration of the tire from the linear acceleration of the bicycle.

The kinematics equations states that

$v = v_0 + at\\4.47 = 0 + 2a\\a = 2.235 ~m/s^2$

where a is the linear acceleration.

The relation with the angular and linear acceleration is

$a = \alpha R$

where R is the radius of the tire. Since it is not given in the question, we will leave it as R.

The angular acceleration of the small pebble is

$\alpha = 2.235/R ~m/s^2$

Answer from: natasniebow

The Solid disk gets to the bottom first. This is because the solid disk has a smaller moment of inertia compared to the hoop. As a result it rolls down the incline faster than the hoop. Although it has a smaller moment of inertia, it has a higher angular velocity.

Explanation:

The steps for the solution to this problem can be found in the attachment below. Dynamics of rigid body has been applied in making a decision as to which body gets to the bottom first. Thank you.

Ahoop and a solid disc are relased from rest atthe top of an incline and allowed to roll down the in

Answer from: Quest

Hi, your answer is not correct. this is a cumulative

Answer from: Quest

Answer: it will take 4.36 seconds to reach her average velocity 10.69 miles/hour.explanation: we have equation of motion $s=ut+\frac{1}{2} at^2$ ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/s $s=0*t+\frac{1}{2} *2.18*t^2$ $s=1.09t^2[/ 2equating equation 1 and 2 [tex]4.75t = 1.09t^2\\ \\ t=4.36 seconds.$ so it will take 4.36 seconds to reach her average velocity 10.69 miles/hour.

Another question on Physics

Physics, 22.06.2019 10:30

Arunner is training for an upcoming marathon by running around a 100-m-diameter circular track at constant speed. let a coordinate system have its origin at the center of the circle with the x-axis pointing east and the y-axis north. the runner starts at (x, y) = (50 m, 0 m) and runs 2.5 times around the track in a clockwise direction.

Answers: 1

Answer

Physics, 22.06.2019 12:30

What’s #16 asking me and how do i solve it?

Answers: 2

Answer

Physics, 22.06.2019 18:00

Find the mass by adding the numbers from the three sliders

Answers: 1

Answer

Physics, 22.06.2019 20:10

Two cannon balls weighing 18 kg and 24 kg are chained together and fired horizontally with a velocity of 165 m/s from the top of a 21-m wall. the chain breaks during the flight of the cannonballs and the 18-kg cannonball strikes the ground at t = 1.5 s, at a distance of 240 m from the foot of the wall, and 7 m to the right of the line of fire. determine the position of the other cannonball at that instant. neglect the resistance of the air.

Answers: 1

Answer

You know the right answer?

Ahoop and a solid disc are relased from rest
atthe top of an incline and allowed to roll down...

Questions