## Supercharge UBQ with

0 attempts

0% avg

UBQ Credits

Verfied Answer
Verfied Explanation 0 likes
0

(a) Calculate the linear speed of the sphere when it reaches the bottom of the incline.

Step Derivation/Formula Reasoning
1 h = L \sin(\theta) Calculate the vertical height h fallen by the sphere using the length of the incline L and the sine of the incline angle \theta.
2 h = 7.0 \sin(35^\circ) Substitute L = 7.0 \, m and \theta = 35^\circ.
3 PE_{\text{top}} = KE_{\text{trans}} + KE_{\text{rot}} Use the conservation of mechanical energy, where potential energy at the top is equal to the sum of transnational kinetic energy and rotational kinetic energy at the bottom.
4 mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 Express the conservation of energy equation in terms of v (linear velocity) and \omega (angular velocity).
5 I = \frac{2}{5}MR^2 Substitute the given moment of inertia for a solid sphere, where I = \frac{2}{5}MR^2.
6 v = R\omega Relation between linear velocity and angular velocity for rolling without slipping.
7 \omega = \frac{v}{R} Rearrange the equation for \omega.
8 mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R})^2 Substitute I and \omega in terms of v and R.
9 mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 Simplify the equation by canceling M and R.
10 mgh = \frac{7}{10}mv^2 Combine like terms.
11 v^2 = \frac{10}{7}gh Isolate v^2.
12 v = \sqrt{\frac{10}{7}gh} Take the square root to find v.
13 v = \sqrt{\frac{10}{7}(9.8)(7.0 \sin(35^\circ))} Substitute the values of g and h.
14 \boxed{v \approx 7.5 \, \text{m/s}} Final answer.

(b) Determine the angular speed of the sphere at the bottom of the incline.

Step Derivation/Formula Reasoning
1 \omega = \frac{v}{R} Use the relation between linear and angular velocities for a rolling object.
2 \omega = \frac{7.5}{0.15} Substitute v = 7.5 \, \text{m/s} and R = 0.15 \, m (converted from cm).
3 \boxed{\omega \approx 50 \, \text{rad/s}} Final answer.

(c) Does the linear speed depend on the radius or mass of the sphere?

Step Analysis Conclusion
1 From the energy conservation equation, the mass m cancelled out and the final expression for v didn’t include the radius R. The linear speed does not depend on the mass or radius of the sphere as both factors were eliminated in deriving v.

(d) Does the angular speed depend on the radius or mass of the sphere?

Step Analysis Conclusion
1 Angular speed \omega was found from v divided by R, but it did not involve mass m. Angular speed does depend on the radius and does not depend on the mass of the sphere.

## Need Help? Ask Phy To Explain This Problem

Phy can also check your working. Just snap a picture!

Simple Chat Box

## See how Others Did on this question | Coming Soon

##### Discussion Threads
###### Login to Discuss
1. v \approx 7.5 \, \text{m/s}
2. \omega \approx 50 \, \text{rad/s}
3. According to the equation derived in part (a) the speed does not depend on the radius or mass of sphere.
4. Angular speed does depend on the radius and does not depend on the mass of the sphere.

## Continue with

By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.

## Share This Question

Enjoying UBQ? Share the 🔗 with friends!

Link Copied!
##### Dark Mode Equation Sheet (Download)
KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g}
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\omega (Angular Velocity)\text{radians per second (rad/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters.

1. Start with the given measurement: \text{5 km}

2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}

3. Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}

4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

## Phy Pro

The most advanced version of Phy. Currently 50% off, for early supporters.

## \$11.99

per month

Billed Monthly. Cancel Anytime.

Trial  –>  Phy Pro

## Error Report

Sign in before submitting feedback.