In the system shown there is no friction between the wheel and its axle. The whe

Question

In the system shown there is no friction between the wheel
and its axle. The wheel is in the form of a solid cylinder with
a radius of 0.2 m. (NOTE: the moment of inertia of a solid
cylinder of mass m and radius r about a central axis is

In the system shown there is no friction between the wheel and its axle. The wheel is in the form of a solid cylinder with a radius of 0.2 m. (NOTE: the moment of inertia of a solid cylinder of mass m and radius r bout a central axis is 1 / 2mr 2.) The block and the wheel each have a mass of one kg. The string is wrapped around the wheel and the system is initially at rest. As the block slides frictionlessly down the plane the string unwraps causing the wheel to rotate. Angle ö = 30o. Find: the angular acceleration of the wheel the angular displacement of the wheel at the end of one second the tension in the string the work done by the torque acting on the wheel the speed of the block after it has moved 0.1 m down the plane using the Principle of Conservation of Energy. Show that motion (kinematics) equations give the same result.A

Explanation / Answer

I = 0.5*m*r^2 = 0.5*1*0.2^2 = 0.02 kg.m^2


let T ia the tension in the string

net force acting on the block.

Fnet = m*g*sin(30) - T

m*a = m*g*sin(30) - T

a = g*sin(30) - T/m

a = 4.9 - T ---(1)( here m =1)

net torque on the wheel

T = r*T

I*alfa = r*T

0.02 *alfa = r*T

0.02*a/r = r*T (since a = r*alfa)

a = r^2*T/0.02

a = 0.2^2*T/0.02

a = 2*T --(2)

from eqns 1 and 2


2*T = 4.9 - T

3*T = 4.9

T = 4.9/3

= 1.6333 N

a = 2*1.6333

= 3.267 m/s^2

a) alfa = a/r = 3.267/0.2 = 16.33 rad/s62

b) theta = 0.5*alfa*t^2 = 0.5*16.33*1^2 = 8.167 radians

c) T = 1.633 N

d) Workdone = T*theta

= r*T*theta

= 0.2*1.6333*8.167

= 2.667 J

e)

v^2 - u^2 = 2*a*s

v = sqrt(2*a*s)

= sqrt(2*3.267*0.1)

= 0.808 m/s

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