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mafiozo [28]
3 years ago
6

In a lab experiment a light flexible string is wrapped around a solid cylinder with mass M = 9.50 kg and radius R. The cylinder

rotates with negligible friction about a stationary horizontal axis. A mass m = 3.00kg is tied at the free end of the string and release and the mass moves downward. Find the magnitude of the acceleration of the object of mass m.

Physics
1 answer:
Lunna [17]3 years ago
5 0

Answer:3.79 m/s^2

Explanation:

Given

Mass M=9.5 kg

m=3 kg

Net Force is equivalent to \sum F=ma

with tension T in the string    

For mass m

mg-T=ma

T=mg-ma--------1

For cylinder

T\cdot R=I\times \alpha

I for solid cylinder is \frac{2}{5}MR^2 , and \alpha =\frac{a}{R}

thus T=\frac{Ma}{2}----2

Substitute the value of T we get

\frac{Ma}{2}=mg-ma

a(\frac{M}{2}+m)=mg

a=\frac{mg}{\frac{M}{2}+m}

a=3.79 m/s^2

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The spacecraft moves at v1 = 0.5c after the initial increment.The equation becomes V2 = V+V1/1+V*V1/c after the second one. 2 V2 = 0.5c+0.50c/1+(0.50c)^2/c^ 2 = 0.80c

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Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface
FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

\tau=I\alpha     (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

\tau=Fd        (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

Fd=I\alpha\\\\I=\frac{Fd}{\alpha}

Finally, you replace the values of all parameters in the previous equation for I:

I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2

The moment of inertia of the door around the hinges is 2 kgm^2

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