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Continue Rotational motion problems and solutions Question -1 Find the Moment of Inertia of a sphere with axis tangent to it? Solution The moment of inertia of the sphere about the axis passing through the center us $I_C=\frac {2}{5}MR^2$ Using Parallel axis theorem, Moment of inertia through the tangent is given by $I_T =I_C + MR^2$ or $I_T= {7}{5}MR^2$ Question -2 A car accelerates uniformly from rest and reaches a Speed of 22.0 m/s in 9.00 s. If the radius of a tire is 29 cm, find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. Solution From $v=u+at$ $a= \frac {v}{t}$ Now Distance covered $s= \frac {1}{2}at^2 = \frac{1}{2}v t$ Number of revolutions= $\frac {s}{2 \pi r}$ Substituting the values Number of revolutions=54.3 revolutions Question 3 A cylinder of Radius 20 cm is rolling along the surface with constant speed of 80 cm/sec a. What is the rotational speed of the cylinder about its axis? b. What is the magnitude and direction of the acceleration of the points on it surface. c. what is the velocity of the point at the top of the cylinder at any moment d. what is the velocity of the point of the cylinder which is in contact with the floor at any moment Solution a. Rotational speed $\omega = \frac {v}{R} = \frac {80}{20} = 4 rad/s$ b.Since there is no acceleration for cylinder , $\alpha = 0$ and we have just radial acceleration $a_r= \omega ^2 r= 3.2 m/s^2$ c. Velocity at the top will be given as $v_T = v + r \omega =1.6 m/s $ d. Velocity at the top will be given as $v_f=0$ Question 4 The angular momentum of a particle moving in a circular orbit with a constant speed remains conserved about a. Any point of the circumference of the circle b. any point inside the circle c.any point outside the circle d. the centre of the circle Solution Answer is (d) Question 5 Two discs of Moment of inertia $I_1$ and $I_2$ about their respective axis rotating with angular velocities $\omega _1$ and $\omega _2$ respectively are brought in face to face with their axis of rotation coincident. The angular velocity of the composite disc will be a.$\frac {I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2}$ b.$\frac {I_1 \omega_1 - I_2 \omega_2}{I_1 - I_2}$ c. $\frac {I_2 \omega_1 + I_1 \omega_2}{I_1 + I_2}$ d. $\frac {I_2 \omega_1 - I_1 \omega_2}{I_1 + I_2}$ Solution Answer is (a) Angular momentum is conserved as no external torque $I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega$ or $\omega =\frac {I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2}$ Question 6 \frac {I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2} Question -6 A cylinder rolls up an inclined plane, reaches some height and then rolls down (without slipping throughout these motions). The directions of the frictional force acting on the cylinder are, a.up the incline while ascending and down the incline while descending. b.up the incline while ascending as well as descending. c.down the incline while ascending and up the incline while descending. d. down the incline while ascending as well as descending. Solution Suppose the inclination of the plane is $\alpha$. The force acting on the cylinder are weight mg and normal reaction N. In the absence of frictional force, the component $mgsin \alpha$ will give an acceleration down the incline and hence contact point will have a tendency to move downward. To oppose this tendency, the frictional force will act upward. This is independent of whether cylinder is moving up or down. Hence answer is (b) Question -7 A disc has a speed of 1200 rpm and it is made to slow down at a uniform rate of 4 rad/s2.Calculate the number of revolution it makes before coming to rest? Solution f=1200 rpm = 20 rotation per sec Initial angular velocity of the disc ($\omega_0$) =$ 2 \pi f= 2 \pi \times 20 =40 \pi $ rad/s Final angular velocity of the disc ($\omega_f$)=0 Angular acceleration (($\alpha$) = -4 rad/s2 Now $\omega_f= \omega _0 + \alpha t$ or $t= \frac {\omega_0}{ \alpha } = 10 \pi$ sec Now $\theta = \omega _0 t + \frac {1}{2} \alpha t^2$ $\theta=200 \pi^2$ rad Therefore, Number of revolutions= $\frac {200 \pi^2}{2 \pi} = 314$ Question 8 A circular disc of mass m and radius r is rolling on a smooth horizontal surface with a constant speed v. Calculate the Kinetic energy of the disc Solution Kinetic Energy of the rolling disc is given by $K = \frac {1}{2}mv^2 + \frac {1}{2} I\omega^2$ Now $I = \frac {1}{2}mr^2$ and $ \omega = \frac {v}{r}$ So, $K = \frac {1}{2}mv^2 + \frac {1}{2} (\frac {1}{2}mr^2) (\frac {v}{r})^2$ $= \frac {3}{4}mv^2 $ Question 9 A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoing circular motion is the x-y plane with centre at O and constant angular speed ?. If the angular momentum of the system, calculated about O and P are denoted by $\vec{L_O}$ and $\vec{L_P}$ respectively, then (A) $\vec{L_O}$ and $\vec{L_P}$ do not vary with time. (B) $\vec{L_O}$ varies with time while $\vec{L_P}$ remains constant. (C) $\vec{L_O}$ remains constant while $\vec{L_P}$ varies with time. (D) $\vec{L_O}$ and $\vec{L_P}$ both vary with time Solution $\vec{L_0}$ is constant both in magnitude and direction $\vec{L_P}$ is constant in magnitude , but direction varies Question 10 A bug of mass m sits on the edge of the circular disc of Radius R and rotating with angular speed $\omega$. How large must be the coefficient of friction between bug and disc be if the bug is not to slip off? Solution For the bug not to slip off, the frictional force $f=\mu mg $ must supplies the centripetal force So, $\mu mg = m \omega^2 R$ or $\mu = \frac {\omega^2 R}{g}$ link to this page by copying the following textRotational motion problems with solutions Also Read When you try to solve problems of Physics in general and of rotational motion in particular, it is important to follow a certain order. Try to be organized when you solve these problems, and you will see how it gives good results. It is worth spending a bit of time on the analysis of a problem before tackling it. When solving a rotational motion problem, follow these steps: Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks! Read carefully the problem statement. Draw a diagram of the physical situation described in the problem. Write down the givens in the problem statement. Identify the elements that constitute the system in motion: whether it is a single body or if on the contrary you must take into account several objects. This step is important because it will allow you to distinguish between the internal and external forces that act on the system. Draw the external forces that act on the system. Remember that forces are physical interactions. Therefore, if a system is supported on another system, a normal force will act on it; if it is close to the surface of the Earth, the weight will act on it, and so on with the rest of the interactions that the system experiences. In this link you can see how the forces exerted by the different types of support are represented. Add a Cartesian coordinate system to the drawing, indicating which is the positive direction of each of the axes. you should also indicate which point you are going to use as an origin to calculate the torque of the external forces acting on the system. If the center of mass of the system moves, write Newton’s second law for it as a vectorial equation, including in the vectorial sum for the different external forces you have identified. Project the resulting equation on the Cartesian axes. Take into account the sign of the different projections. If the system rotates, write Newton’s second law for rotation in a vectorial form. Calculate the torques of the different external forces with respect to the point you have chosen as origin. Project the resulting equation on the Cartesian axes. Take into account the sign of the different projections. Solve the system of equations that you have obtained after projecting the different equations on the axes. In this way you will be able to determine the magnitudes asked in the problem. Do not forget to include the units in your results. Review the problem and check that the results you have obtained make sense. On the following pages you will find some rotational motion problems with solutions. Try to do them before looking at the solution. The post Rotational motion problems and solutions appeared first on YouPhysics Rotational Motion Exam1 and Problem Solutions 1. An object, attached to a 0,5m string, does 4 rotation in one second. Find a) Period b) Tangential velocity c) Angular velocity of the object. a) If the object does 4 rotation in one second, its frequency becomes; f=4s-1 T=1/f=1/4s b) Tangential velocity of the object; V=2.π.f.r V=2.3.4.0,5 V=12m/s c) Angular velocity of the object ω=2.π.f=2.3.4=24radian/s 2. Find the relation between tangential and angular velocities of points X, Y and Z. X and Y rotate together, so if X does one rotation then Y also does one rotation. On the contrary, if Y does one revolution, Z does two revolutions. Angular velocities of the X, Y and Z are; ωX=ωY=ωZ/2 3. An object hanged on a rope L=0,5m, does rotational motion. If the angle between rope and vertical is 370, find the tangential velocity of the object. (g=10m/s2, cos370=0,8, sin370=0,6) Free body diagram of system is given below; Horizontal component of tension on the rope makes object rotate. TX=mV2/r, TY=m.g Radius of the motion path is; r=L.sin370=0,5.0,6=0,3m tan370=TX/TY 3/4=mV2/r/m.g 3/4=V2/g.r V=3/2m/s 4. An object having mass m does rotational motion. Its angular velocity is ω and radius of motion path is r. Find kinetic energy of the object in terms of r, ω, and m. EK=1/2m.V2 V=ω.r EK=1/2m(ω.r)2 EK=mω2.r2/2 5. Stone having mass 0,5kg rotates in horizontal. It is hanged on 1m rope. If the tension on the rope is 80 N, find the frequency of the motion. Fnet=80N=m.ω2.r 80=m.4.π2.f2.r 80=0,5.4.32.f2.1 f=2s-1 Tags: Torque 1. A beam 140 cm in length. There are three forces acts on the beam, F1 = 20 N, F2 = 10 N, and F3 = 40 N with direction and position as shown in the figure below. What is the torque causes the beam rotates about the center of mass of the beam? Known : The center of mass located at the center of the beam. Length of beam (l) = 140 cm = 1.4 meters Force 1 (F1) = 20 N, the lever arm 1 (l1) = 70 cm = 0.7 meters Force 2 (F2) = 10 N, the lever arm 2 (l2) = 100 cm – 70 cm = 30 cm = 0.3 meters Force 3 (F3) = 40 N, the lever arm 3 (l3) = 70 cm = 0.7 meters Wanted : The magnitude of torque Solution : The torque 1 rotates beam clockwise, so assigned a negative sign to the torque 1. τ1 = F1 l1 = (20 N)(0.7 m) = -14 N m The torque 2 rotates beam counterclockwise, so assigned a positive sign to the torque 2. τ2 = F2 l2 = (10 N)(0.3 m) = 3 N m The torque 3 rotates beam clockwise, so assigned a positive sign to the torque 3. τ3 = F3 l3 = (40 N)(0.7 m) = -28 N m The net torque : Στ = -14 Nm + 3 Nm – 28 Nm = – 42 Nm + 3 Nm = -39 Nm The magnitude of the torque is 39 N m. The direction of rotation of the beam clockwise, so assigned a negative sign. Read : Graphical of linear motion – problems and solutions 2. What is the net torque acts on the beam The axis of rotation at point D. (sin 53o = 0.8) Known : The axis of rotation at point D F1 = 10 N and l1 = r1 sin θ = (40 cm)(sin 53o) = (0.4 m)(0.8) = 0.32 meters F2 = 10√2 N and l2 = r2 sin θ = (20 cm)(sin 45o) = (0.2 m)(0.5√2) = 0.1√2 meters F3 = 20 N and l3 = r1 sin θ = (10 cm) (sin 90o) = (0.1 m)(1) = 0.1 meters Wanted : The net torque Solution : τ1 = F1 l1 = (10 N)(0.32 m) = 3.2 Nm (The torque 1 rotates beam counterclockwise so we assign positive sign to the torque 1) τ2 = F2 l2 = (10√2 N)( 0.1√2 m) = -2 Nm (The torque 2 rotates beam clockwise so we assign negative sign to the torque 2) τ3 = F2 l2 = (20 N)(0.1 m) = 2 Nm (The torque 3 rotates beam counterclockwise so we assign positive sign to the torque 3) The net torque : Στ = τ1 – τ1 + τ3 Στ = 3.2 Nm – 2 Nm + 2 Nm Στ = 3.2 Nm Read : Radioactivity – problems and solutions 3. What is the net torque if the axis of rotation at point D. (sin 53o = 0.8) Known : The axis of rotation at point D. Distance between F1 and the axis of rotation (rAD) = 40 cm = 0.4 m Distance between F2 and the axis of rotation (rBD) = 20 cm = 0.2 m Distance between F3 and the axis of rotation (rCD) = 10 cm = 0.1 m F1 = 10 Newton F2 = 10√2 Newton F3 = 20 Newton Sin 53o = 0.8 Wanted : The net torque Solution : The moment of the force 1 Στ1 = (F1)(rAD sin 53o) = (10 N) (0.4 m)(0.8) = 3.2 N.m (The torque 1 rotates beam counterclockwise so we assign positive sign to the torque 1) The moment of the force 2 Στ2 = (F2)(rBD sin 45o) = (10√2 N)(0.2 m)(0.5√2) = -2 N.m (The torque 2 rotates beam clockwise so we assign negative sign to the torque 2) The moment of the force 3 Στ3 = (F3)(rCD sin 90o) = (20 N)(0.1 m)(1) = 2 N.m (The torque 2 rotates beam counterclockwise so we assign positive sign to the torque 3) The net torque : Στ = Στ1 + Στ2 + Στ3 Στ = 3.2 – 2 + 2 Στ = 3.2 Newton meter Read : Determine the position of an object in projectile motion The moment of inertia 4. Length of wire = 12 m, l1 = 4 m. Ignore wire’s mass. What is the moment of inertia of the system. Known : Mass of A (mA) = 0.2 kg Mass of B (mB) = 0.6 kg Distance between A and the axis of rotation (rA) = 4 meters Distance between B and the axis of rotation (rB) = 12 – 4 = 8 meters Wanted : The moment of inertia of the system Solution : The moment of inertia of A IA = (mA)(rA2) = (0.2)(4)2 = (0.2)(16) = 3.2 kg m2 The moment of inertia of B IB = (mB)(rB2) = (0.6)(8)2 = (0.6)(64) = 38.4 kg m2 The moment of inertia of the system : I = IA + IB = 3.2 + 38.4 = 41.6 kg m2 Read : Pressure of solids – problems and solutions Rotational dynamics 5. A 6-N force is applied to a cord wrapped around a pulley of mass M = 5 kg and radius R = 20 cm. What is the angular acceleration of the pulley. The pulley is a uniform solid cylinder. Known : Force (F) = 6 Newton Mass (M) = 5 kg Radius (R) = 20 cm = 20/100 m = 0.2 m Wanted : Angular acceleration (α) Solution : The moment of the force : τ = F R = (6 Newton)(0.2 meters) = 1.2 N m The moment of inertia for solid cylinder : I = 1/2 M R2 I = 1/2 (5 kg)(0.2 m)2 I = 1/2 (5 kg)(0.04 m2) I = 1/2 (0.2) I = 0.1 kg m2. The angular acceleration : τ = I α α = τ / I = 1.2 / 0.1 = 12 rad s-2 Read : Optical instrument contact lenses – problems and solutions 6. A block of mass = 4 kg hanging from a cord wrapped around a pulley of mass = 8 kg and radius R = 10 cm. Acceleration due to gravity is 10 ms-2 . What is the linear acceleration of the block? The pulley is a uniform solid cylinder. Known : Mass of pulley (m) = 8 kg Radius of pulley (r) = 10 cm = 0.1 m Mass of block (m) = 4 kg Acceleration due to gravity (g) = 10 m/s2 Weight (w) = m g = (4 kg)(10 m/s2) = 40 kg m/s2 = 40 Newton Wanted : The free fall acceleration of the block Solution : The moment of inertia of the solid cylinder : I = 1/2 M R2 = 1/2 (8 kg)(0.1 m)2 = (4 kg)(0.01 m2) = 0.04 kg m2 The moment of the force : τ = F r = (40 N)(0.1 m) = 4 Nm The angular acceleration : Στ = I α 4 = 0.04 α α = 4 / 0.04 = 100 The linear acceleration : a = r α = (0.1)(100) = 10 m/s2 7. A block with mass of m hanging from a cord wrapped around a pulley. If the free fall acceleration of the block is a m/s2, what is the moment of inertia of the pulley.. Known : weight = w = m g Lever arm = R The angular acceleration = α The free fall acceleration of the block = a ms-2 Wanted: The moment of inertia of the pulley (I) Solution : The connection between the linear acceleration and the angular acceleration : a = R α α = a / R The moment of inertia : τ = I α I = τ : α = τ : a / R = τ (R / a) = τ R a-1 Read : Springs in series and parallel – problems and solutions The angular momentum 8. A 0.2-gram particle moves in a circle at a constant speed of 10 m/s. The radius of the circle is 3 cm. What is the angular momentum of the particle? Known : Mass of particle (m) = 0.2 gram = 2 x 10-4 kg Angular speed (ω) = 10 rad s-1 Radius (r) = 3 cm = 3 x 10-2 meters Wanted : The angular momentum of the particle Solution : The equation of the angular momentum : L = I ω I = the angular momentum, I = the moment of inertia, ω = the angular speed The moment of inertia (for particle) : I = m r2 = (2 x 10-4 )(3 x 10-2)2 = (2 x 10-4 ) (9 x 10-4) = 18 x 10-8 The angular momentum : L = I ω = (18 x 10-8)(10 rad s-1) = 18 x 10-7 kg m2 s-1 korupigog.pdf 95825829462.pdf pitching in business pdf parenikevu.pdf 48366628836.pdf 96509010821.pdf 86369171995.pdf apex launcher frp agamben karman pdf 1607d94fa5022c---69262601825.pdf winrar mac os 15212164821.pdf kill a watt how to use cual es el proposito de una entrevista de trabajo 160a3fd3974afa---ludilivoravixak.pdf 16074416672e85---1010305168.pdf gesobimavovakarib.pdf zoo tycoon 2 indir 50 shades of grey full movie online xaruzito.pdf androidrock voice changer apk download neologism examples pdf benefits of organisational structure тесты по технологии для мальчиков 7 класс
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