A circular saw blade 22.0 cm in diameter starts from rest. In 3.10 s, it acceler

Question

A circular saw blade 22.0 cm in diameter starts from rest. In 3.10 s, it accelerates with constant angular acceleration to an angular velocity of +7000 rev/min. Find the angular acceleration m the blade spins up in rad/s^2. Through how many radians has it turned during this timespan? The website for one manufacturer lists "Maximum RPM" for saw blades. What is the linear speed of a tooth at the edge of a saw blade of diameter 40.0 cm operating at its maximum RPM of 4000? Express your answer in m/s. (Understanding that answer should help you to understand why larger diameter saw blades have lower maximum RPM recommendations.) Four flat objects (circular ring, circular disk, square disk, and square ring) have the same mass M and the same outer dimension (circular objects have diameters of 2R and squares have sides of 2R). The small circle at the center of each figure represents the axis of rotation for these objects. This axis of rotation passes through the center of mass and a perpendicular to the plane of the objects Rank, from greatest to least, the moments of inertia of the objects. Greatest. 1_____2______3______4_______Least OR, The moments of inertia of all the objects will be the same OR, We cannot determine the ranking for moments of inertia of the objects

Explanation / Answer

a)

V = V(0) + a*t

where

V = 7000 rev/min = 733 rad/sec

a = V / t = 733 rad/s / 3.10 s = 236.45 rad / s^2

b)

s = ut + (1/2)at^2
here it starts from rest so u = 0

s = angle travelled..
a is the angular acceleration
t is the time

s = 1/2*236.45*3.1^2

s = 1136 radians.

Not able to find out c and d.

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