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Question: A rope of negligible mass is wound around a hollow cylinder of mass \[{\mathbf{3kg}}\] and radius \[...

A rope of negligible mass is wound around a hollow cylinder of mass 3kg{\mathbf{3kg}} and radius 40cm.{\mathbf{40cm}}. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30N30N? What is the linear acceleration of the rope? Assume that there is no slipping.

Explanation

Solution

To solve this question,you must have an idea about moment of inertia, torque, angular acceleration and must have knowledge that torque is directly proportional to moment of inertia and angular acceleration.

Complete step by step answer:
Moment of inertia: it is the inertia of a rotating body to its rotation. It is a rotating body's resistance to angular acceleration or deceleration, equal to the product of the mass and the square of its perpendicular distance from the axis of rotation.
Torque: Torque is a measure of the force that can cause an object to rotate about an axis.
Angular acceleration: The angular acceleration is the time rate of change of the angular velocity and is usually designated by α and expressed in radians per second squared.
In question the given values are;
Mass of the hollow cylinder (m) = 3kg\left( m \right){\text{ }} = {\text{ }}3kg
Radius of the hollow cylinder (r) = 40cm =0.40m\left( r \right){\text{ }} = {\text{ }}40cm{\text{ }} = 0.40m
Applied force (F) = 30N\left( F \right){\text{ }} = {\text{ }}30N
The moment of inertia of the hollow cylinder about its geometric axis:
I=mr2I = m{r^2}
=3×0.42kgm2= 3 \times {0.4^2}kg{m^2}
=1.26kgm2= 1.26kg{m^2}
The Torque developed is given by:
τ=rF\tau = rF
=0.4×30Nm= 0.4 \times 30Nm
=12Nm= 12Nm
We know that the relation between the torque and angular acceleration is:

\tau = I\alpha \\\ \alpha = \dfrac{\tau }{I} \\\ \alpha = \dfrac{{1.26kg{m^2}}}{{12Nm}} \\\ \\\ {\alpha = 25\dfrac{{rad}}{{{s^{ 2}}}}} \end{array}$$ Where $$\alpha = $$ angular acceleration $$I = $$ moment of inertia $$\tau = $$ torque Thus linear acceleration: $$a = r\alpha $$ $$ = 0.4 \times 25\dfrac{m}{{{s^2}}}$$ $$ = 10\dfrac{m}{{{s^2}}}$$ **Note:** Sometimes students get confused in the term acceleration and angular acceleration so be kept in the mind that whenever we talk about rotational motion then there is always angular acceleration and incase of linear motion there is acceleration.