Question

A rope is wound around a uniform solid cylinder of mass
6 kg and radius 80 cm. The rope is pulled with a force F
such that angular acceleration of the cylinder is 10
rad/s2. The value of F is​

Answer 1

Answer:

it's correct

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Answer 2

Given:

A rope is wound around a uniform solid cylinder of mass 6 kg and radius 80 cm. The rope is pulled with a force F such that angular acceleration of the cylinder is 10 rad/s².

To find:

Value of F ?

Calculation:

The rotational analogue of force is called Torque, it can be represented as follows:

$\therefore \: | \tau| = F \times r \times \sin( \theta)$

• F is force , r is perpendicular distance of force and axis , $\theta$ is angle between them.

• Now $\tau$ can also be represented as the product of Moment of inertia and angular acceleration.

$\implies \: I \times \alpha = F \times r \times \sin( \theta)$

Putting necessary values:

$\implies \: I \times \alpha = F \times r \times \sin( {90}^{ \circ} )$

$\implies \: \bigg( \dfrac{m {r}^{2} }{2} \bigg) \times \alpha = F \times r \times \sin( {90}^{ \circ} )$

$\implies \: \bigg( \dfrac{m {r}^{2} }{2} \bigg) \times \alpha = F \times r$

$\implies \: F = \dfrac{m \times r \times \alpha }{2}$

$\implies \: F = \dfrac{6 \times \frac{80}{100} \times 10 }{2}$

$\implies \: F = 6 \times \dfrac{80}{100} \times 5$

$\implies \: F = 24 \: N$

So, the value of force is 24 N.