Answers. Best Answer: This question uses one of the definitions of torque: T_net = I * alpha where T is the torque, I is the moment of inertia, and alpha is the angular acceleration. This is akin to Newton's second law, but for rotational motion. T_net = 4.0x10^-5 * 150 = 0.006 N-m.
Oct 16, 2008· Physics question-grinding wheel? A small grinding wheel has a moment of inertia of 4.0×10−5 kg*m^2. What net torque must be applied to the wheel
Answer to A small grinding wheel has a moment of inertia of 4.0*10^-5 kg*m2 . What net torque must be applied to the wheel for its...
Answer to A small grinding wheel has a moment of inertia of 4.0... Skip Navigation. Chegg home. Books. Study. Textbook Solutions Expert Q&A. A Small Grinding Wheel Has A Moment Of Inertia Of 4.0. This problem has been solved! See the answer. A small grinding wheel has a moment of inertia of 4.0.
Answer (1 of 1): The formula of net torque is as follows, and by just putting the values in the formula, net torque can be calculated. Net Torque = Moment of inertia x Angular acceleration Net Torque = 4.0×10−5 x 150Net Torque = 600 ×10−5Net Torque = 6.0×10−3 N.m
A grinding wheel is a uniform cylinder with a radius of 8.50 cm and a mass of 0.380 kg. Calculate (a) its moment of inertia about its center, and (b) the applied torque needed to accelerate it from rest to 1750 rpm in 5.00 s. Take into account a frictional torque that has been measured to slow down the wheel
Oct 28, 2010· A grinding wheel is initially at rest. A constant external torque of 52.5 N· m is applied to the wheel for 18.4 s, giving the wheel an angular speed of 605 rev/min. The external torque is then removed, and the wheel comes to rest 101 s later. Find the moment of inertia of the wheel. 2. Homework Equations Torque = I * a 3. The Attempt at a Solution
40 J. a solid sphere with a mass of 4.0 kg and a radius of 0.10 m starts from rest at the top of a ramp inclined at 30 degrees and rolls to the bottom. the upper end of the ramp is 1.0
rearranged to solve for the required moment of inertia for the steel shaft: != Pa2(a+b) 3E" ST = (60)(12)2(12+20) 3(29 x 106)(.00174) =18264 in4 3. The moment of inertia equation for a round shaft is rearranged to solve for diameter: d= 64 I! 4= 64(18264)! 4 =2.47 in 4. Alternatively, the equation given below can be used to solve for the new shaft size directly. d new =d
Physics Exam #2. A hydraulic press for compacting powdered samples has a large cylinder which is 10.0 cm in diameter, and a small cylinder with a diameter of 2.0 cm. A lever is attached to the small cylinder. There sample, which is placed on the large cylinder, has an area of 4.0 cm^2.
The wearing down of rock elements by friction due to water, wind or ice is also known as grinding.It also means the reducing of ore into very tiny particles by means of pressure or collision. Diverse types of grinders are used in the processing place to obtain the desired measurement.Grinding is a derogatory term used in computer gaming to portray the process of engaging in repetitive gameplay.
Answer to A small grinding wheel has a moment of inertia of 4.0105 kg*2. What net torque must be applied to the wheel for its angular acceleration to be 150
we knew I, we could solve for !. The moment of inertia depends on the shape of the object, its distribution of mass, and the axis through which it rotates. Since this requires some calculus to derive from scratch, we must merely look up the moment of inertia for our objects. For a solid disk rotating through its center, the moment of inertia is I= 1 2 MR 2
Calculate the moment of inertia of the door about the hinge line. 2(e) A solid disk (mass M = 3.00 kg and radius R = 20.0 cm) is hung from the wall by means of a metal pin through the hole, and used as a pendulum. Calculate the moment of inertia of the disk about the pin (= the axis of the rotation).
A grinding wheel is a uniform cylinder with a radius of 8.25 cm and a mass of 750. g. Calculate (a) its moment of inertia about its center, and (b) the applied torque needed to accelerate it from rest to 2500. rpm in 9.50 s if it is known to slow down from 1250 rpm to rest in exactly 1 minute.
(b) A 1 inch US Standard Washer has inner radius r. i = 13:5 mm and an outer radius r. o = 31:0 mm. The washer is approximately d= 4:0 mm thick. The density of the washer is ˆ= 7:8 10. 3. kg m. 3. Calculate the moment of inertia of the washer about an axis that is perpendicular to the plane of the washer and passes through its center of mass.
rotational inertia of an object plays the same role as ordinary mass for simple motion • For a given amount of torque applied to an object, its rotational inertia determines its rotational acceleration Æthe smaller the rotational inertia, the bigger the rotational acceleration Big rotational inertia Small rotational inertia Same torque
A grinding wheel is a uniform cylinder with a radius of 8.70 cm and a mass of 0.400 kg. Part A) Calculate its moment of inertia about its center. Express your answer with the appropriate units
negligible mass passing over a pulley of radius 0.220 m and moment of inertia I. The block on the frictionless incline is moving with a constant acceleration of magnitude a = 1.40 m/s2. (Let m 1 = 13.5 kg, m 2 = 18.0 kg, and = 37:0 .) From this information, we wish to nd the moment of inertia of the pulley.
Insight: This speed corresponds to about 1.4 mi/h, half the normal walking speed of an adult. The larger wheels on adult bicycles allow for higher linear speeds for the same angular speed of the driving wheel. Kinetic Energy of a Rotating Wheel. A typical ten-pound car
A grinding wheel is a uniform cylinder with a radius of 8.90 cm and a mass of 0.560 kg.? (a) Calculate its moment of inertia about its center. kg·m2 (b) Calculate the applied torque needed to accelerate it from rest to 1200 rpm in 2.00 s if it is known to slow down from 1200 rpm to rest in 52.0 s. If you believe your intellectual property
Recall that the moment of inertia for a point particle is I = mr 2. The torque applied perpendicularly to the point mass in Figure 10.37 is therefore $$\tau = I \alpha \ldotp$$ The torque on the particle is equal to the moment of inertia about the rotation axis times the angular acceleration. We can generalize this equation to a rigid body rotating about a fixed axis.