1. Show that the moment of inertia for a solid cylinder with a radius R, mass M,

Question

1. Show that the moment of inertia for a solid cylinder with a radius R, mass M, and length L, rotating about its central axis is given by: I=1/2MR^2

2. With axle and spokes of negligible mass and a thin rim, a certain bicycle wheel has a radius of 0.350 m and weighs 37.0 N; it can turn on its axle with negligible friction. A man holds the wheel above his head with the axle vertical while he stands on a turntable that is free to rotate without friction; the wheel rotates clockwise, as seen from above, with an angular speed of 57.7 rad/s, and the turntable is initially at rest. The rotational inertia of wheel + man + turntable about the common axis of rotation is 2.10 kgm^2.The man's free hand suddenly stops the rotation of the wheel (relative to the turntable). Determined the resulting (a) angular speed and (b) direction of rotation of the system.

3. A block weighing 14.0 N, which can slide without friction on an incline at angle of 40, is connected to the top of the incline by a massless spring of unstretched length 0.450 m and spring constant 120 N/m. (a) How far from the top of the incline is the block's equilibrium point? (b) If the block is pulled slightly down the incline and released, what is the period of the resulting oscillations?

4. Consider drilling a tunnel between two opposite points on the Earths surface passing through the center of the Earth, and assume that the tunnel is frictionless and that the Earth has uniform density. (a) If we place a subway car inside the tunnel at a distance r from the center of the Earth, what will be the force of gravity acting on the subway car from Earth? (b) Prove that the subway car placed in the tunnel will execute simple harmonic motion and nd the angular frequency of the motion in the tunnel. 1. Show that the moment of inertia for a solid cylinder with a radius R, mass M, and length L, rotating about its central axis is given by: I=1/2MR^2

2. With axle and spokes of negligible mass and a thin rim, a certain bicycle wheel has a radius of 0.350 m and weighs 37.0 N; it can turn on its axle with negligible friction. A man holds the wheel above his head with the axle vertical while he stands on a turntable that is free to rotate without friction; the wheel rotates clockwise, as seen from above, with an angular speed of 57.7 rad/s, and the turntable is initially at rest. The rotational inertia of wheel + man + turntable about the common axis of rotation is 2.10 kgm^2.The man's free hand suddenly stops the rotation of the wheel (relative to the turntable). Determined the resulting (a) angular speed and (b) direction of rotation of the system. 2. With axle and spokes of negligible mass and a thin rim, a certain bicycle wheel has a radius of 0.350 m and weighs 37.0 N; it can turn on its axle with negligible friction. A man holds the wheel above his head with the axle vertical while he stands on a turntable that is free to rotate without friction; the wheel rotates clockwise, as seen from above, with an angular speed of 57.7 rad/s, and the turntable is initially at rest. The rotational inertia of wheel + man + turntable about the common axis of rotation is 2.10 kgm^2.The man's free hand suddenly stops the rotation of the wheel (relative to the turntable). Determined the resulting (a) angular speed and (b) direction of rotation of the system.

3. A block weighing 14.0 N, which can slide without friction on an incline at angle of 40, is connected to the top of the incline by a massless spring of unstretched length 0.450 m and spring constant 120 N/m. (a) How far from the top of the incline is the block's equilibrium point? (b) If the block is pulled slightly down the incline and released, what is the period of the resulting oscillations? 3. A block weighing 14.0 N, which can slide without friction on an incline at angle of 40, is connected to the top of the incline by a massless spring of unstretched length 0.450 m and spring constant 120 N/m. (a) How far from the top of the incline is the block's equilibrium point? (b) If the block is pulled slightly down the incline and released, what is the period of the resulting oscillations?

4. Consider drilling a tunnel between two opposite points on the Earths surface passing through the center of the Earth, and assume that the tunnel is frictionless and that the Earth has uniform density. (a) If we place a subway car inside the tunnel at a distance r from the center of the Earth, what will be the force of gravity acting on the subway car from Earth? (b) Prove that the subway car placed in the tunnel will execute simple harmonic motion and nd the angular frequency of the motion in the tunnel.
4. Consider drilling a tunnel between two opposite points on the Earths surface passing through the center of the Earth, and assume that the tunnel is frictionless and that the Earth has uniform density. (a) If we place a subway car inside the tunnel at a distance r from the center of the Earth, what will be the force of gravity acting on the subway car from Earth? (b) Prove that the subway car placed in the tunnel will execute simple harmonic motion and nd the angular frequency of the motion in the tunnel.

Explanation / Answer

1)

Take an infinitesimal cylindrical section of length L, width dr, located at radius r. Assume density of cylinder = rho.

infinitesimal mass dm = rho*(2*pi*r dr)*L

MOI = Integral (r = 0 to R) dm * r^2

= Integral (r = 0 to R) rho*(2*pi*r dr)*L * r^2

= 2*pi*L*rho * Integral (r = 0 to R) r^3 dr

= 2*pi*L*rho * R^4 /4

But mass M = rho*Volume

M = rho*(pi*R^2 *L)

Thus, MOI = 1/2 *MR^2

2)

W = mg

m = 37 / 9.81 = 3.772 kg

MOI of wheel = mr^2 = 3.772*0.35^2 = 0.462 kg-m^2

Total angular momentum of system = I*w = 0.462*57.7 = 26.66 kg-m^2 /s

Final MOI = I_total*w_new

26.66 = 2.1*w_new

w_new = 12.695 rad/s

Time period T = 2*pi/w = 2*3.14/12.695 = 0.4947 s

b) Directon = clockwise when view from top. (since w_new = positive)

3)

a)

F = k*x0 = mg*Sin theta

120*x0 = 14*sin40

x0 = 0.075 m

Distance of equilib point = 0.45 + 0.075 = 0.525 m

b)

mx'' = mg*sin theta - k(x + x0)

But mg*sin theta = k*x0

Thus, m*x'' = -kx

mx'' + kx = 0

Assume, x = a * sin (wt) then x'' = -x*w^2

w = sqrt (k/m)

w = sqrt (120 / (14/9.81))

w = 9.17 rad/s

4)

a)

The force on the mass is strictly radial toward the earth center and is given by the

law of universal gravitation as-

F_r = G*m*M_r / r^2

Now GM = g*R^2

assume a uniform density earth such that the mass Mr lying below r relates to the total earth mass M as

Mr / M = (r / R)^3

Thus,

Force of gravity = mg*(r/R)..........where R = Radius of earth.

We have made use of the fact that the attractive force at radial distance r inside a uniform density sphere with radius R involves only the mass lying below r.

b)

If we now formulate the equation of motion for the mass m in the tunnel, we get

m x'' = -mg x/R

applying the initial conditions that x(0)=x0 and dx(0)/dt=0, one gets the very simple solution

x(t) = x_0 Cos (sqrt (g/R) *t)

This result shows that the mass m is undergoing simple harmonic motion with the angular frequency ?=(2?/?)=sqrt(mg/R)

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