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Mars2501 [29]
3 years ago
6

A block (mass = 74.0 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, wh

ere M = 7.9 is the mass of the pulley and R=1.4 is its radius ), as the drawing shows. Initially the pulley is prevented from rotating and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of r=R=1.4 m during the block's descent, and the distance of the block at initial position to the floor is 7.0 m. What is the angular velocity in round per minute when the block drops to the floor? Use g = 10 m/s2.
Physics
1 answer:
katen-ka-za [31]3 years ago
4 0

Answer:

51.2m/s

Explanation:

Gravity force acting on the block:

F = mg = 74 * 10 = 740 N

Torque acting on the pulley:

T = FR = 740 * 1.4 = 1036 Nm

Moments of inertia of the solid pulley

I = \frac{1}{2}MR^2 = 0.5*7.9*1.4^2 = 7.742 kgm^2

The angular acceleration of the pulley after release:

\alpha = \frac{T}{I} = \frac{1036}{7.742} = 133.8 rad/s^2

The linear acceleration or the rope

a = \alpha R = 133.8 * 1.4 = 187.34 m/s^2

The time it takes for the block to reach the floor:

s = \frac{1}{2}at^2

t^2 = \frac{2s}{a} = \frac{2 * 7}{187.34} = 0.0747

t = \sqrt{0.0747} = 0.273 s

The final velocity at the floor would be

v = at = 187.34 * 0.273 = 51.2 m/s

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Answer:

=\frac{1/3}{5/6} = 0.4

Explanation:

Moment of inertia of given shell= \frac{2}{3} MR^2

where

M represent sphere mass

R -sphere radius

we know linear speed is given as v = r\omega

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(a) -39.4^{\circ}

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

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To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m

The initial total momentum along the x-direction as

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The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

mu = 6.69 m\\u = 6.69 m/s

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