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Tutorial 6

16-70

. If the slider block

C

is moving at $v_{C}=3 \mathrm{~m} /...

Sep 30, 2024

Solution by Steps

step 1

Given that the slider block

C

is moving at

v_C = 3 \, \mathrm{m/s}

, we can find the angular velocity of the crank

AB

using the relationship between linear and angular velocity. The length of crank

AB

0.5 \, \mathrm{m}

, so the angular velocity

\omega_{AB}

can be calculated as:

\omega_{AB} = \frac{v_C}{L_{AB}} = \frac{3}{0.5} = 6 \, \mathrm{rad/s}.

step 2

To find the angular velocity of the rod

BC

, we need to consider the geometry of the system. The angle between crank

AB

and the horizontal is

60^\circ

, and the angle between rod

BC

and the horizontal is

45^\circ

. The relationship between the angular velocities can be expressed as:

\omega_{BC} = \omega_{AB} \cdot \frac{L_{AB}}{L_{BC}} \cdot \cos(60^\circ) = 6 \cdot \frac{0.5}{1} \cdot \cos(60^\circ) = 6 \cdot 0.5 \cdot 0.5 = 1.5 \, \mathrm{rad/s}.

step 3

Therefore, the angular velocities of the crank

AB

and the rod

BC

at the instant shown are:

\omega_{AB} = 6 \, \mathrm{rad/s}, \quad \omega_{BC} = 1.5 \, \mathrm{rad/s}.

Answer

Angular velocity of crank

AB

6 \, \mathrm{rad/s}

and angular velocity of rod

BC

1.5 \, \mathrm{rad/s}

Key Concept

The relationship between linear and angular velocity in mechanical systems.

Explanation

The angular velocities are derived from the linear velocity of the slider block and the geometry of the system, demonstrating how motion is transmitted through connected components.

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Solution by Steps

step 1

For the spool rolling without slipping, the relationship between the linear velocity

v

of the cable and the velocity of the center

O

of the spool can be expressed as:

v = R \cdot \omega,

where

R

is the radius of the spool and

\omega

is the angular velocity of the spool

step 2

Since the spool rolls without slipping, the linear velocity of the center

O

is equal to the velocity of the cable being pulled. Therefore, we can express the velocity of the center

O

as:

v_O = v.

step 3

Thus, the velocity of the center

O

of the spool when the cable is pulled to the right with a velocity

v

is:

v_O = v.

Answer

The velocity of the center

O

of the spool is equal to the velocity

v

of the cable being pulled.

Key Concept

The relationship between linear velocity and angular velocity in rolling motion.

Explanation

The center of the spool moves with the same velocity as the cable due to the no-slip condition, illustrating the connection between linear and angular motion in rolling objects.