A pulley of moment of inertia 2.0 kg-m^2 is mounted on a wall as shown in the following figure.

A pulley of moment of inertia 2.0 kg-m^2 is mounted on a...

A pulley of moment of inertia $2.0 \mathrm{kg}-\mathrm{m}^{2}$ is mounted on a wall as shown in the following figure. Light strings are wrapped around two circumferences of the pulley and weights are attached. What are (a) the angular acceleration of the pulley and (b) the linear acceleration of the weights? Assume the following data: $$r_{1}=50 \mathrm{cm}, \quad r_{2}=20 \mathrm{cm}, \quad m_{1}=1.0 \mathrm{kg}, \quad m_{2}=2.0 \mathrm{kg}$$

Key Concepts

Relationship Between Angular and Linear Acceleration

In rotating systems, there is a direct relationship between angular acceleration and the linear acceleration of points along the rim of a rotating body. This relationship is given by the equation a = r × ?, where r is the radius from the axis of rotation and ? is the angular acceleration. It allows for the translation of rotational motion into linear motion.

Moment of Inertia

This is the property of a body that quantifies its resistance to changes in its rotational motion about an axis. The larger the moment of inertia, the harder it is to cause any given angular acceleration. It depends on the mass distribution relative to the axis of rotation and plays a crucial role in the dynamics of rotational systems.

Torque

Torque is the rotational equivalent of force and is defined as the product of a force and the lever arm (the perpendicular distance from the axis of rotation to the line of action of the force). It is the mechanism through which forces cause changes in rotational motion, and the net torque on an object will cause it to accelerate rotationally.

Angular Acceleration

Angular acceleration is the rate of change of angular velocity over time. In rotational dynamics, angular acceleration is directly proportional to the net applied torque and inversely proportional to the moment of inertia, according to Newton's second law for rotation.

Newton's Second Law for Rotation

This law states that the net external torque acting on a system is equal to the product of the system's moment of inertia and the angular acceleration (?_net = I·?). It forms the foundational principle for analyzing problems involving rotational motion, paralleling the role of F = m·a in linear dynamics.

Transcript

00:01 Question number 92 in this question there is a pulley with different radiae the mass m2 is hanging like this this is m2 g and here t 2 tension here is hanging a mass m1 with m1 g as weight this is tension t1 this is also tension t1 this tension is acting on the pulley creating a torque on the pulley the pulley is turning in this direction with angular acceleration alpha the linear acceleration of mass m2 is a2 and the linear acceleration of mass m1 is a1 so now let us write the equations for mass m1 and 2 for mass m1 we can write m1 m1 g minus t1 should be equal to m1 a1 similarly for mass m2 we can write t 2 minus m2 g should be equal to m2 a2 now the net torque on the pulley is equal to t 1 minus t2 into i2 and this net torque should be equal to movement of inertia of the pulley into its angular acceleration so now substituting the values of t1 and t2 t1 is m1 g minus m1 a 1 into r1 minus t2 value is m2 a 2 and plus m2 g into r2 and this should be equal to movement of inertia into angular exhalation now expanding this we can write m1 g r1 minus m1 and the linear acceleration a1 can also be written as r1 alpha so it will become r1 square into alpha minus m2 a 2 r2 r2 this a 2 r2 can be written as alpha r2 square because linear acceleration is equal to r2 alpha so instead of a 2 we can write alpha into r2 square minus m2 g r2 r2 this is equal to i alpha the relation between angular acceleration and linear acceleration is a is equal to r alpha now substituting the values m1 is 1 into 9 .8 into 0 .5 minus 1 1 into alpha into 0 .5 square minus m2 which is 2 into alpha into 0 .2 square minus 2 into 9 .8 into r2 which is 0 .2 meter this should be equal to movement of inertia 2 into alpha solving this we will get the angular acceleration of the pulley alpha as 0 .4206 radian per second square.

05:29 In the part b, we have to find the linear acceleration of mass m1, a1 is equal to r1 alpha.

05:44 This is equal to r1 is 0 .5 into alpha 0 .4 -206...

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