Physics: A cord is wrapped around the rim of a solid uniform wheel 0.280 m in radius and of mass 7.80 kg. A steady horizontal pull of 34.0 N to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. Required: a. Compute the angular acceleration of the wheel b. Compute the acceleration of the part of the cord that has already been pulled off the wheel. c. Find the magnitude of the force that the axle

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A cord is wrapped around the rim of a solid uniform

Orabah

04.08.2021

Solution :

Given :

Mass, m = 7.80 kg

Radius, r = 0.280 m

Force, F = 34 N

a). St = I x a

$34 \times r = 7.8 r^2 \times \frac{a}{2}$

$a=\frac{68}{7.8 \times 0.280}$

$=31.13 \ rad/s^2$

b). The acceleration of the part of the chord is

= 31.13 x 0.280

$=8.7164 \ m/s^2$

c). Magnitude of the force is = $\sqrt{34^2 + (7.8 \times 9.81)^2}$

= 83.73 N

d). The direction of force is = arctan (7.8 x 9.81)/34

= 2.25

e). The answers in part (c) and (d) will change when the pull of force is given in the upward direction instead of giving in the horizontal direction

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