8. When a grinder is switched on, a 1.75 kg, 22.0 cm diameter grinding wheel (disk) uniformly accelerates to 350 revolutions per minute (rpm) in 5.42 seconds. What is the angular acceleration of the grinding wheel? rad/s How many revolutions does the wheel make during this 5.42 seconds? What is the torque supplied by the motor to the grinding wheel during this time? N•m
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- In the figure below, a wheel of radius 0.25 m is mounted on a frictionless horizontal axle. A massless cord is wrapped around the wheel and attached to a 2.0 kg box that slides on a frictionless surface inclined at angle e = 25° with the horizontal. The box accelerates down the surface at 2.1 m/s?. What is the rotational inertia of the wheel about the axle? |kg • m2A rotating wheel requires 5.00 s to rotate 29.0 revolutions. Its angular velocity at the end of the 5.00-s interval is 96.0 rad/s. What is the constant angular acceleration (in rad/s) of the wheel? rad/s22.5 3. A playground merry-go-round of radius 1.5 meters. The child applies a constant force of F, =50 Newtons to the rim. There is a frictional torque of 35 Nm opposing the rotation. The graph shows a plot of the resulting angular velocity of the merry-go-round as a function of time. 2 1.5 0.5 Fo view from above time (s) a. What is the torque due to the force F, (take ccw as positive)? b. What is the moment of inertia of the merry-go-round? @ (rad/sec)
- A fan blade is rotating with a constant angular acceleration of +11.2 rad/s2. At what point on the blade, as measured from the axis of rotation, does the magnitude of the tangential acceleration equal that of the acceleration due to gravity? iA solid 0.5950 kg ball rolls without slipping down a track toward a vertical loop of radius ?=0.6550 m. What minimum translational speed ?min must the ball have when it is a height ?=1.008 m above the bottom of the loop in order to complete the loop without falling off the track? Assume that the radius of the ball itself is much smaller than the loop radius ?. Use ?=9.810 m/s2 for the acceleration due to gravity.4. A skater has a moment of inertia of 105.0 kg.m² when his arms are outstretched and a moment of inertia of 64.0 kg.m² when his arms are tucked in close to his chest. If he starts to spin at an angular speed of 80.0 rpm (revolutions per minute) with his arms outstretched, what will his angular speed be when they are tucked in? rpm
- A wheel is rotating counter clockwise initially at an angular speed of 2.50 rad/s. Itdeceleratesover a 4.0-s interval at a rate of 0.140 rad/s2. a)What is its angular speed after this 7.0-s interval? b)What is the angular displacement? c) How long would it take for it to stop?A skater spins about a fixed point on the ice. She begins with her arms extended and an initial angular velocity wo. She then pulls her arms in to her body. After her arms are pulled to her body, she spins with an angular velocity final ,wf . Throughout the time she is spinning, no external forces are acting in the horizontal plane. How do the magnitudes of the initial and final angular velocities compare? OA. Wo > Wf OB. WO = wf w0 < wf OE. ооо ооA person 1.84 m tall swings their leg from the hip at 2.16 radians/s. Calculate the linear tangential velocity (v) at their knee to 2 decimal places. page Answer: nere to search R 6 H N O P Next page 6°C Rain