MECHANICS Lecture notes for Phys 111 Abstract
... beats per min. How many gallons of blood does the heart pump in 1 year? ( 1 gallon= 3800 cm3 ). ...
... beats per min. How many gallons of blood does the heart pump in 1 year? ( 1 gallon= 3800 cm3 ). ...
Exam 2
... At the highest point in its trajectory, its speed is 100 m/s. If air resistance is ignored, the initial velocity had a horizontal component of a. 100 m/s. b. 100/sin(25o) m/s. c. 100 sin(25o) m/s. d. 0 m/s. e. 100 cos(25o) m/s. 2. A large cannon is fired from ground level over level ground at an ang ...
... At the highest point in its trajectory, its speed is 100 m/s. If air resistance is ignored, the initial velocity had a horizontal component of a. 100 m/s. b. 100/sin(25o) m/s. c. 100 sin(25o) m/s. d. 0 m/s. e. 100 cos(25o) m/s. 2. A large cannon is fired from ground level over level ground at an ang ...
11-2 Vector Cross Product
... 11-1 Angular Momentum—Objects Rotating About a Fixed Axis The rotational analog of linear momentum is angular momentum, L: Then the rotational analog of Newton’s second law is: This form of Newton’s second law is valid even if I is not constant. ...
... 11-1 Angular Momentum—Objects Rotating About a Fixed Axis The rotational analog of linear momentum is angular momentum, L: Then the rotational analog of Newton’s second law is: This form of Newton’s second law is valid even if I is not constant. ...
Ch 12 Notes – Teacher2 - Mona Shores Public Schools
... Regarding Newton’s 2nd Law of Motion • The acceleration of an object is always in the same direction as the net force. • In using Newton’s second law, it is helpful to realize that the units N/kg and m/s2 are equivalent • Newton’s second law also applies when a net force acts in the direction opposi ...
... Regarding Newton’s 2nd Law of Motion • The acceleration of an object is always in the same direction as the net force. • In using Newton’s second law, it is helpful to realize that the units N/kg and m/s2 are equivalent • Newton’s second law also applies when a net force acts in the direction opposi ...
Chapter 12
... the center of gravity of the object If g is uniform over the object, then the center of gravity of the object coincides with its center of mass If the object is homogeneous and symmetrical, the center of gravity coincides with its geometric center ...
... the center of gravity of the object If g is uniform over the object, then the center of gravity of the object coincides with its center of mass If the object is homogeneous and symmetrical, the center of gravity coincides with its geometric center ...
Chapter 12
... the center of gravity of the object If g is uniform over the object, then the center of gravity of the object coincides with its center of mass If the object is homogeneous and symmetrical, the center of gravity coincides with its geometric center ...
... the center of gravity of the object If g is uniform over the object, then the center of gravity of the object coincides with its center of mass If the object is homogeneous and symmetrical, the center of gravity coincides with its geometric center ...
rotary motion - GEOCITIES.ws
... A 7.5 kg bowling ball (Irolling = 7/5 m R2 ) with a radius of 12 cm is at the top of a ramp 5.0 m long and inclined at 30.00. a) Find the torque acting on the ball. b) Find its rotational inertia. ...
... A 7.5 kg bowling ball (Irolling = 7/5 m R2 ) with a radius of 12 cm is at the top of a ramp 5.0 m long and inclined at 30.00. a) Find the torque acting on the ball. b) Find its rotational inertia. ...
Section 2 Forces, Energy and Power
... Newton’s 2nd law of motion states that the acceleration of an object: varies directly as the unbalanced force applied if the mass is constant varies inversely as the mass if the unbalanced force is constant. These can be combined to give a α F m a = kF where k is a constant m kF = ma The unit of for ...
... Newton’s 2nd law of motion states that the acceleration of an object: varies directly as the unbalanced force applied if the mass is constant varies inversely as the mass if the unbalanced force is constant. These can be combined to give a α F m a = kF where k is a constant m kF = ma The unit of for ...
Ch 08 B1 QFD.cwk (WP)
... The object shown above consists of two disks that are glued together so that they rotate freely about an axis that passes through their centers and is perpendicular to the page. If the net torque on the object is zero then it remains at rest. What is the magnitude of the force, labeled X, that is re ...
... The object shown above consists of two disks that are glued together so that they rotate freely about an axis that passes through their centers and is perpendicular to the page. If the net torque on the object is zero then it remains at rest. What is the magnitude of the force, labeled X, that is re ...
Glider and Pulley
... Connect the other end of the thread to the mass hanger and hold the glider at the end of the track. Open the EasySense software package, click on Timing, then select Raw Times. Hold the glider at the other end and add a small mass of 5 g to the mass hanger. Once you have clicked Start, let g ...
... Connect the other end of the thread to the mass hanger and hold the glider at the end of the track. Open the EasySense software package, click on Timing, then select Raw Times. Hold the glider at the other end and add a small mass of 5 g to the mass hanger. Once you have clicked Start, let g ...
ppt document
... St = dL/dt Just like SF = dp/dt leads to Conservation of momentum if no external forces are present, so St = dL/dt leads to Conservation of Angular momentum if no external torques are present. Note: p = mv, and L = r p = r mv = r m vq = r m r = mr2 = I . ...
... St = dL/dt Just like SF = dp/dt leads to Conservation of momentum if no external forces are present, so St = dL/dt leads to Conservation of Angular momentum if no external torques are present. Note: p = mv, and L = r p = r mv = r m vq = r m r = mr2 = I . ...
Chapter 8 Momentum, Impulse and Collisions
... dt i where ⃗p is a new physical quantity known as momentum. In this course we define it as ⃗p ≡ m⃗v. ...
... dt i where ⃗p is a new physical quantity known as momentum. In this course we define it as ⃗p ≡ m⃗v. ...
Center of mass
In physics, the center of mass of a distribution of mass in space is the unique point where the weighted relative position of the distributed mass sums to zero or the point where if a force is applied causes it to move in direction of force without rotation. The distribution of mass is balanced around the center of mass and the average of the weighted position coordinates of the distributed mass defines its coordinates. Calculations in mechanics are often simplified when formulated with respect to the center of mass.In the case of a single rigid body, the center of mass is fixed in relation to the body, and if the body has uniform density, it will be located at the centroid. The center of mass may be located outside the physical body, as is sometimes the case for hollow or open-shaped objects, such as a horseshoe. In the case of a distribution of separate bodies, such as the planets of the Solar System, the center of mass may not correspond to the position of any individual member of the system.The center of mass is a useful reference point for calculations in mechanics that involve masses distributed in space, such as the linear and angular momentum of planetary bodies and rigid body dynamics. In orbital mechanics, the equations of motion of planets are formulated as point masses located at the centers of mass. The center of mass frame is an inertial frame in which the center of mass of a system is at rest with respect to the origin of the coordinate system.