4. A wheel in the shape of a uniform disk ( I = 1/2 MR^2 )...

4. A wheel in the shape of a uniform disk ( I = 1/2 MR^2 ) has radius 0.0800 m and mass 4.00 kg . The wheel is mounted on a horizontal frictionless axle that passes through its center. A light rope is wrapped around the wheel and a block of mass 5.00 kg is suspended from the free end of the rope. The system is released from rest. What is the angular speed of the rotating wheel after the block has descended 6.00 m? (a) 101 rad/s (b) 115 rad/s (c) 136 rad/s (d) 143 rad/s (e) none of the above answers 5. A uniform disk with radius 0.200 m is mounted on a vertical frictionless axle at its center. A light rope is wrapped around the disk. The disk is initially at rest and then starting at t = 0 a constant force F = 8.00 N is applied to the free end of the rope and the disk turns through 3.00 rev in 2.00 s. What is the moment of inertia of the disk for an axis along the axle? (a) 0.17 kg•m^2 (b) 0.34 kg•m^2 (c) 0.85 kg•m^2 (d) 1.07 kg•m^2 (e) 9.42 kg•m^2 (f) none of the above answers

Transcript

00:01 In this question we are given here a wheel of the shape we are given here a wheel in the shape of the uniform dix we are given here a shape of uniform digs we are given there is the uniform and moment of finish here i which is equal to we are given here 1 divided by 2 m rm am a little by r square the radius we are talking about so r which is equal we can write 0 .0 80 80 meter and the mass we are given here so small m which is equal to we can write 4 kilograms so mass is 4 kilograms.

00:32 The wheel is mounted on a horizontal fix in the xl and the pass into the center.

00:36 So suppose we have this wheel and here this is using the string using something that we are here put this block and block block mass is m.

00:51 Now here we understand that the light rope we are using light rope.

00:56 Here we are using the light rope and the wrapped around the arm the wheel and the block mass so mass of the block which is equal to 5 kudang this is the mass of the wheel now suspended from the free end of the rope and the system is released from the rest so here system it is from the rest so it is zero meter per second now what is the angular speed of the rotating wheel after the block has a descending to 6 meter so here d which is equal to 6mmd descending to 6 meter so we can write you want to an angular speed speed omega that we want to find so let's hold this question and finding the answer here so here first understand that the rope is not slipping and applying the energy so a e total we can wait total energy which is equal you can write kinetic energy plus we can wait the potential energy we know the velocity is zero so this values all zero now now here we can write the e total which is equal to potential energy only so we get which is equal you can wait the mass of the block multiple g multiple h mg h this is the energy now when the object is released to a total energy after the descending six meter after descending six meters so you can write e total which is equal we can write the kinetic energy of translation the kinetic energy of translation plus we can write the kinetic energy of rotation thus we can write the change in the potential energy which is substitute value so we can write 1 divided by 2 mv by v squared plus we can write 1 divided 8 to a multiple by omega square plus we can write m g here we can write m g substitute value into here we just simplify and we can write here 1 divided by 2 mv plus we can write m d divided by d and here this value is multiplied by v square now plus we can write mv there is mb here v g or d however h2 so this is the equation here which is equal to we can write the equation of e total energy value of total energy now here this is the equation number 2 here this is the equation number 2 and we have equation number 1 this is our equation number 1 and this is height is h 1 this is the equation over 1 now we know the both equations is same so we can write it we just simplify the both equation and we can write here solving 1 and 2 equation so then we can write a v square which is equal to we substitute value here so 5 multiplied by 9 .8 multiplied by 6 divided by 5 we can write 1 divided by 2 5 plus we can write 4 divided by 4 it is we just simplifying we can write your answer is so we score which is equal to values 84 .09 and now if we just simply find we can write your answer is 9 .17 meter per second so this is the value of velocity v here now let's write the value of omega so we can write value of omega which is equal we can write v divided by capital r so value of v here the value of v is 9 .17 divided by value of r which is equal to 0 .0 which is equal to we just simplify and we can write here answer is 1 .44 .625 and answer is in the radiant per second so this is the value of the omega so here the radius is already we are given as 0.

04:36 0 .8 .0.

04:38 Now, after simplification, we can write the value of omega is approximately equal to 1115 gradient per second.

04:47 So this is the value of angular speed...

Question

Please give Ace some feedback